Question

A block of mass 3.7 kg, which has an initial velocity of 7.9 m/s at time t = 0, slides on a horizontal surface. Calculate the work that must be done on the block to bring it to rest. Answer in units of J.

Answer 1

Answer:

115.6 J

Explanation:

$m$ = mass of the block = 3.7 kg

$v_{f}$ = final speed of the block = 0 ms⁻¹

$v_{o}$ = initial speed of the block = 7.9 ms⁻¹

$W$ = work done on the block

Using work done - change in kinetic energy theorem, we have

$W = (0.5) m (v_{o}^{2} - v_{f}^{2})\\W = (0.5) (3.7) (7.9^{2} - 0^{2})\\W = 115.6 J$

Answer 2

115.46 J work must be done on the block to bring it to rest.

Explanation:

From law of conservation of energy, neglecting the friction, we know that:

Work done to stop the body = Kinetic energy lost by the body

Work done to stop the body = 1/2 mv²

Work done to stop the body = 1/2 (3.7 kg) (7.9 m/s)²

Work done to stop the body = 115.46 J