A block of mass 2.0 kg is released from rest on a slope. It travels 7.0 m down the slope and falls a vertical distance of 3.0 m. The block experiences a frictional force parallel to the slope of 5.0 N.

Question 3

A block of mass 2.0 kg is released from rest on a slope. It travels 7.0 m down the slope and falls a vertical distance of 3.0 m. The block experiences a frictional force parallel to the slope of 5.0 N.

What is the speed of the block after falling this distance?

A 4.9 m s^-1                        B 6.6 m s^-1                        C 8.6 m s^-1                        D 10.1 m s^-1

Reference: Past Exam Paper – June 2011 Paper 11 & 13 Q15

Solution:

Answer: A.

At the top of the slope, the block has gravitational potential energy (GPE). As the block falls, some GPE is converted into kinetic energy of the block and work done due the frictional force.

ΔGPE = KE + Work done against friction

The loss in height Δh of the block is 3.0 m.

ΔGPE = mgΔh = 2.0 × 9.81 × 3.0 = 58.86 N

Frictional force = 5.0 N

Distance moved against friction = 7.0 m

Work done against friction = F × d = 5.0 × 7.0 = 35.0 J

ΔGPE = KE + Work done against friction

58.86 = KE + 35.0

KE of block at bottom = 58.86 – 35.0 = 23.86 J

½ mv² = 23.86

Speed after falling = 4.9 m s^-1