Question

Suppose a 5.00 kg box starts from rest and slides 2.50 m down a ramp. The ramp makes an angle of 30.0° with the horizontal. The coefficient of kinetic friction between the box and the ramp is µk = 0.436. Determine the work done on the box by the friction force between the box and the ramp.

Answer 1

The workdone by the friction force is
`W_F= 46.25J`

Step-by-step explanation:

From the question we are told that

The mass of the box is
`m_b = 5.00kg`

The distance it slides
`d = 2.50m`

The angle made by the with the horizontal is
`\theta = 30.0^o`

The coefficient of kinetic friction is
`\mu_k = 0.436`

Workdone by frictional force is mathematically represented as

`W_F = Frictional \ force * distance traveled`

Now
`Frictional \ Force = mg \mu_k cos(\theta)`

the cos
`\theta` is because the force is acting in horizontal direction and from the the question we can deduce that to resolve the force acting on the box horizontally we need to use cos
`\theta`

Substituting we have

`W_F = m\ * g\ * d \ * \mu_k cos (\theta)`

Substituting value we have that

`W_F = 5 * 9.8 * 2.5 * 0.436 * cos (30.0^o)`

`W_F= 46.25J`