Question

A crate of mass 9.6 kg is pulled up a rough incline with an initial speed of 1.44 m/s. The pulling force is 104 N parallel to the incline, which makes an angle of 20.3° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.02 m.
(a) How much work is done by the gravitational force on the crate?
(b) Determine the increase in internal energy of the crate–incline system owing to friction.
(c) How much work is done by the 104-N force on the crate?
(d) What is the change in kinetic energy of the crate?
(e) What is the speed of the crate after being pulled 5.02 m?

Answer 1

Work and Kinetic Energy

We need to find the work done, kinetic energy, increase in internal energy and speed of the crate

Solution:

Given

m= mass = 9.6 kg

u = initial speed = 1.44 m/s.

F =pulling force = 104 N

$\theta = angle = 20.3^o$

$u_k =$coefficient of kinetic friction =0.4

d= distance = 5.02 m

(a).

work done by gravitational force

$$= -m \times g \times d\times sin \space \theta\\$$

$$= -9.6 \times 9.8 \times 5.02\times sin \space 20.3^o$$

work done by gravitational force

$$=- 163.85$$

(b).

Increase in internal energy of the crate

$$= -u_k \times m \times g \times cos \space\theta \times d$$

$$= - 0. 4 \times 9.6 \times 9.8 \times cos \space 20.3^o \times 5.02$$

$$= -177 .17 \space J$$

(c).

work is done by the 104-N force on the crate

$$=F \times d$$

$$= (104 N) \times (5.02 \space m) = 522.08 \space J$$

(d)

Using work-energy theorem,

change in kinetic energy of the crate = the work done

$$= - 163.85 - 177.17 + 522.08 =181.06 \space J$$

(e)

speed of the crate after being pulled 5.02 m

$$= \sqrt {\frac {2 \Delta K.E} {m} + u^2}$$

$$= \sqrt {\frac {2 (181.06 J)} {9.6kg} + (1.44)^2}$$

$$= \sqrt {39.79} = 6.3 m/s$$