Unit 3: Work & Energy Review

Authored by Kevin Martilla

Physics

12th Grade

NGSS covered

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13 questions

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MATCH QUESTION

1 min • 1 pt

Match the following terms with the correct equation.

$mgh$

Kinetic Energy

$\frac{1}{2}mv^2$

Conservation of Energy

$\frac{1}{2}mv^2+mgh$

Work-Energy Theorem

$\frac{1}{2}mv_i^2+mgh_i+F_{nc}d=\frac{1}{2}mv_f^2+mgh_f$

Gravitational Potential Energy

$W=\Delta KE$

Mechanical Energy

MATCH QUESTION

1 min • 1 pt

Match the following

energy of motion

Mechanical Energy

the ability to do work

Energy

the sum of kinetic & potential energy

Kinetic Energy

energy from position/shape/ config.

Potential Energy

Tags

NGSS.MS-PS3-1

NGSS.MS-PS3-2

NGSS.MS-PS3-5

MULTIPLE SELECT QUESTION

2 mins • 1 pt

Work requires two things:
(Select two answers)

Any force acting on an object

Displacement of an object

A force with a component in the same or opposite direction of motion

A force perpendicular to the displacement of an object

A displaced object moving with a constant velocity

Tags

NGSS.HS-PS3-1

MULTIPLE SELECT QUESTION

45 sec • 1 pt

The following describes the Work-Energy Theorem:
(Select all that apply)

The net work done on an object is equal to its change in kinetic energy

$W_{net}=\Delta KE$

$W_{net}=\frac{1}{2}m\left(v_f^2-v_i^2\right)$

$F_{net}\cdot d=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

What equation do you use to find the velocity of an object given its kinetic energy?

$v=\sqrt[]{\frac{2KE}{m}}$

$v=\frac{2KE}{m}$

$v=\frac{1}{2}mv^2$

$v=\frac{KE}{mgh}$

Tags

NGSS.HS-PS3-1

MULTIPLE SELECT QUESTION

45 sec • 1 pt

What is power?
(Select all that apply)

The rate at which work is done

$\frac{Work}{time}$

$\frac{F\cdot d}{t}$

Measured in $\frac{Joules}{\sec ond}$

Measured in Watts (W)

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

A force, F=240N, is applied to a sled of mass, m=57 kg, pushing at an angle, θ=30^o to the horizontal. The force of friction is F_f=30 N. If the sled moves a distance 90 m along a flat surface, what is the work done by the applied force on the box?

18,700 J

21,600 J

18,900 J

50,300 J

Answer explanation

Given: $F_A=240\ N;\ \theta=30\degree;\ d=90m$
$W=F_A\ d\cos\theta\ =\left(240N\right)\left(90m\right)\cos30\degree=18,706J$

Tags

NGSS.HS-PS3-1

NGSS.HS-PS2-1

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

A force, F=240N, is applied to a sled of mass, m=57 kg, pushing at an angle, θ=30^o to the horizontal. The force of friction is F_f=30 N. If the sled moves a distance 90 m along a flat surface, what is the work done by the friction force on the box?

-2,700 J

2,700 J

-25,100 J

50,300 J

Answer explanation

Given: $F_f=30\ N;\ \theta=180\degree;\ d=90m$
$W=F_f\ d\cos\theta\ =\left(30N\right)\left(90m\right)\cos180\degree=-2,700J$

Tags

NGSS.HS-PS3-1

NGSS.HS-PS3-2

MULTIPLE CHOICE QUESTION

1 min • 1 pt

A force, F=240N, is applied to a sled of mass, m=57 kg, pushing at an angle, θ=30^o to the horizontal. The force of friction is F_f=30 N. If the sled moves a distance 90 m along a flat surface, what is the total work done on the box?

16,000 J

2,700 J

1,8700 J

0 J

Answer explanation

Given: $W_{F_A}=18,706\ J;\ W_{F_f}=-2,700\ J$
$W_{tot}=W_{F_A}+W_{F_f}=16,006\ J$

Tags

NGSS.HS-PS3-1

10.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

What is the net work done on an elevator that is accelerated upward at 4.2 m/s² through a distance of 20 m? The total mass of the elevator and its 8 passengers is 1,200 kg .

235,200 J

100,800 J

336,000 J

0 J

-100,800 J

Answer explanation

Net Work is caused by the Net Force applied over a distance.
$W_{net}=F_{net}d=mad=\left(1,200kg\right)\left(4.2\ \frac{m}{s^2}\right)\left(20m\right)=100,800\ J$

Tags

NGSS.HS-PS2-1

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Unit 3: Work & Energy Review

Match the following terms with the correct equation.

Match the following

Work requires two things:(Select two answers)

The following describes the Work-Energy Theorem:(Select all that apply)

What equation do you use to find the velocity of an object given its kinetic energy?

What is power?(Select all that apply)

A force, F=240N, is applied to a sled of mass, m=57 kg, pushing at an angle, θ=30o to the horizontal. The force of friction is Ff=30 N. If the sled moves a distance 90 m along a flat surface, what is the work done by the applied force on the box?

Given: FA=240 N; θ=30°; d=90mF_A=240\ N;\ \theta=30\degree;\ d=90mFA​=240 N; θ=30°; d=90m W=FA dcos⁡θ =(240N)(90m)cos⁡30°=18,706JW=F_A\ d\cos\theta\ =\left(240N\right)\left(90m\right)\cos30\degree=18,706JW=FA​ dcosθ =(240N)(90m)cos30°=18,706J

A force, F=240N, is applied to a sled of mass, m=57 kg, pushing at an angle, θ=30o to the horizontal. The force of friction is Ff=30 N. If the sled moves a distance 90 m along a flat surface, what is the work done by the friction force on the box?

Given: Ff=30 N; θ=180°; d=90mF_f=30\ N;\ \theta=180\degree;\ d=90mFf​=30 N; θ=180°; d=90m W=Ff dcos⁡θ =(30N)(90m)cos⁡180°=−2,700JW=F_f\ d\cos\theta\ =\left(30N\right)\left(90m\right)\cos180\degree=-2,700JW=Ff​ dcosθ =(30N)(90m)cos180°=−2,700J

A force, F=240N, is applied to a sled of mass, m=57 kg, pushing at an angle, θ=30o to the horizontal. The force of friction is Ff=30 N. If the sled moves a distance 90 m along a flat surface, what is the total work done on the box?

Given: WFA=18,706 J; WFf=−2,700 JW_{F_A}=18,706\ J;\ W_{F_f}=-2,700\ JWFA​​=18,706 J; WFf​​=−2,700 J Wtot=WFA+WFf=16,006 JW_{tot}=W_{F_A}+W_{F_f}=16,006\ JWtot​=WFA​​+WFf​​=16,006 J

What is the net work done on an elevator that is accelerated upward at 4.2 m/s2 through a distance of 20 m? The total mass of the elevator and its 8 passengers is 1,200 kg .

Net Work is caused by the Net Force applied over a distance. Wnet=Fnetd=mad=(1,200kg)(4.2 ms2)(20m)=100,800 JW_{net}=F_{net}d=mad=\left(1,200kg\right)\left(4.2\ \frac{m}{s^2}\right)\left(20m\right)=100,800\ JWnet​=Fnet​d=mad=(1,200kg)(4.2 s2m​)(20m)=100,800 J

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Work requires two things:
(Select two answers)

The following describes the Work-Energy Theorem:
(Select all that apply)

What is power?
(Select all that apply)

A force, F=240N, is applied to a sled of mass, m=57 kg, pushing at an angle, θ=30^o to the horizontal. The force of friction is F_f=30 N. If the sled moves a distance 90 m along a flat surface, what is the work done by the applied force on the box?

Given: $F_A=240\ N;\ \theta=30\degree;\ d=90m$
$W=F_A\ d\cos\theta\ =\left(240N\right)\left(90m\right)\cos30\degree=18,706J$

A force, F=240N, is applied to a sled of mass, m=57 kg, pushing at an angle, θ=30^o to the horizontal. The force of friction is F_f=30 N. If the sled moves a distance 90 m along a flat surface, what is the work done by the friction force on the box?

Given: $F_f=30\ N;\ \theta=180\degree;\ d=90m$
$W=F_f\ d\cos\theta\ =\left(30N\right)\left(90m\right)\cos180\degree=-2,700J$

A force, F=240N, is applied to a sled of mass, m=57 kg, pushing at an angle, θ=30^o to the horizontal. The force of friction is F_f=30 N. If the sled moves a distance 90 m along a flat surface, what is the total work done on the box?

Given: $W_{F_A}=18,706\ J;\ W_{F_f}=-2,700\ J$
$W_{tot}=W_{F_A}+W_{F_f}=16,006\ J$

What is the net work done on an elevator that is accelerated upward at 4.2 m/s² through a distance of 20 m? The total mass of the elevator and its 8 passengers is 1,200 kg .

Net Work is caused by the Net Force applied over a distance.
$W_{net}=F_{net}d=mad=\left(1,200kg\right)\left(4.2\ \frac{m}{s^2}\right)\left(20m\right)=100,800\ J$