WEP Session 1

A constant force F is applied to move a box along the floor at constant speed against frictional force of 6.0 N. The box is moved a displacement of 5.0 m along the floor. What is work done by friction?

• 30 J

• -30 J

• 0.0 J

• 15.0 J

Answer explanation

Work done by frictional force = F x s x cos 180 = 6.0 (5.0)(-1.0) =-30 J

A constant force F is applied to move a box along the floor at constant speed against frictional force of 6.0 N. The box moved a distance of 5.0 m along the floor. What is the work done by F?

• 30 J

• -30 J

• 0.0 J

• 15 J

Answer explanation

Work done by F = F.s cos 0

= 6.0 (5.0)

= 30 J

A constant F is applied to move a box along the floor at a constant speed against a frictional force of 6.0 N. The box moved a distance of 5.0 m along the floor.

What is the work done by its weight if the weight of the box is 4.0 N?

• 20 J

• -20 J

• 0.0 J

• 30 J

Answer explanation

W.D by weight = F.scos90 =4.0 (5.0) cos 90 = 0.0 J

Force F is applied to an object as it is moved through a displacement, x, from x =0 to 6.0 m on a level ground. If KEi of the object is 2.0 J, given that there is no friction, the final KE is _ J

• 11.5 J

• 15.5 J

• 13.5J

• 0.0 J

Answer explanation

Total WD by F = area under the F s graph = 1/2 (3+6)(3)=13.5 J

Total WD goes in increase in KE

13.5 J = KEf- KEi

KEf = 13.5 + 2.0 = 15.5 J

A 1.0 kg block moves along the x-axis with no friction acting against its motion.

It passes x=0 with a KE of 2.0 J. Varying force F is acting on it, as shown in the graph.

What is the KE when it reaches 5.0 m?

• 7.0 J

• 5.0 J

• 1.0 J

• 0.0 J

Answer explanation

Area under the graph = WD by F = 2(2.0) - 1/2(1)(2.0) = 3.0 J

WD by F = Increase in KE

3.0 J = KEf-KEi

3.0 J = KEf - 2.0 J

KEf = 5.0 J