Question

A machine pulls a 46 kg trunk 3.0 m up a 42o ramp at constant velocity, with the machine's force on the trunk directed parallel to the ramp. The coefficient of kinetic friction between the trunk and the ramp is 0.36. What are (a) the work done on the trunk by the machine's force and (b) the increase in thermal energy of the trunk and the ramp?

Answer 1

Answer:

the answers are

W = 1271.256 J

$E_{th}$= 361.81 J

Explanation:

The mass of the trunk (m) = 46kg

angle between the ramp and the horizontal (θ) = 42°

The trunk is pulled at a displacement (d) up the ramp = 3m

The coefficient of kinetic friction between the trunk and the ramp = 0.36

The trunk is moving with a constant velocity therefore the net force on it is zero, therefore the force required to move the trunk must be equal to the summation of forces opposing the trunk

The two forces opposing the trunk are

1. the gravitational force directed down the ramp  $F_{g}$ and
2. the frictional force between the ramp and the trunk $F_{f}$

We have to calculate the machine's force which is equal to sum of the

$F =F_{g} + F_{f}$

$F_{g}$ = $mgsin$θ

$F_{f} =$μ× N

N = mgcosθ

$F_{f} =$ μ$mgcos$θ

F = mg (sinθ + μcosθ)

F = 46× 9.8 (sin42 + 0.36×cos42)

F= 450.8 (0.67 +0.27)

F = 450.8 × 0.94

F = 423.752N

to calculate the workdone on the trunk by the machine force

The workdone on the trunk is  = W = F × dcosΘ

Θ = 0° because the trunk is directed parallel to the ramp

W =423.752× 3 cos 0

W = 423.752 ×3

W = 1271.256 J

(b)  the increase in thermal energy of the trunk and the ramp

Friction converts mechanical energy into thermal energy, so multiplying the frictional force with the distance gives the thermal energy generated by the trunk

$E_{th} =F_{f}$ × d

= μ$mgcos$θ ×d

= 0.36 ×46×9.8×cos42×3

= 361.81 J

Answer 2

Answer:

(a) W = 1267J

(b) Wother = 362J

Explanation:

In this problem 4 forces act on the trunk; the weight mg of the trunk, F the force of the machine in the trunk and the kinetic frictional force Fk and the reaction force of the ramp surface on the trunk R.

R = mgCosθ

Fk = μR = μmgCosθ

W = mg

The workdone W = F×d

Since the force of the machine is directed parallel to the ramp

Given m = 46kg, θ = 42° d = 3.0m

By resolving the forces acting on the trunk into components, the component of the weight acting parallel to the ramp is mgSin42°

Now by Newton's first law the sum of forces acting on the trunk any direction is zero as the trunk moves with constant velocity.

So summing forces acting on the trunk parallel to the surface of the ramp

F – mgsin42° –Fk = 0

Fk = Kinetic friction = μR = μmgCos42°

F – mgsin42° – μmgCos42° = 0

F = mg(Sin42° + 0.36Cos42°)

F = 46×9.8×(Sin42° + 0.36Cos42°)

F = 422.25N

Workdone = 422.25×3.00 = 1267J

(b) Increase in thermal energy of the trunk is as a result of the workdone on the trunk by the kinetic frictional force.

Wother = Fk×d = μmgCos42°×d

Wother = 0.36×46×9.8Cos42°×3.00

Wother = 362J