Question

A forklift raises a 90.5 kg crate 1.80 m. (a) Showing all your work and using unity conversion ratios, calculate the work done by the forklift on the crane, in units of kJ. (b) If it takes 12.3 seconds to lift the crate, calculate the useful power supplied to the crate in kilowatts.

Answer 1

(a) 1.629 kJ

(b) 0.132 kW

Step-by-step explanation:

The workdone (W) on a body is the product of the force (F) acting on the body and the displacement (s) caused by the force. This can be written as;

W = F x s ----------------(i)

Where;

F = m x a [From Newton's law] ------------------(ii)

m = mass of the body

a = acceleration of the body

Since the direction of motion is upwards, the acceleration acting on the body is the one due to gravity.

i.e a = g = 10m/s²

From the question, the mass of the crate is 90.5kg.

Substitute these values into equation (ii) to get the force acting on the body as follows;

F = 90.5 x 10

F = 905N

From the question;

The displacement (s) of the crate is 1.80m

Substitute the values of the force F and the displacement s into equation (i) as follows;

W = 905 x 1.8

W = 1629 J

Convert the result to kJ by dividing by 1000 as follows;

W = (1629/1000) kJ

W = 1.629 kJ

Therefore, the work done on the crate is 1.629 kJ

(b) Power (P) is the ratio of work done (W) to the time taken (t) to do the work. i.e;

P = W / t -----------------(ii)

From the question;

t = 12.3s

Also;

W = 1.629 kJ [as calculated in a]

Substitute these values into equation(ii) as follows;

P = 1.629 / 12.3

P = 0.132 kW

Therefore the useful power supplied to the crate is 0.132 kW

Answer 2

(a) Work done is 1.59642 kJ

(b) Useful power supplied = 0.1298 kW

Step-by-step explanation:

(a) Work done = mass of crate × acceleration due to gravity × distance = 90.5 kg × 9.8 m/s^2 × 1.8 m = 1596.42 kgm^2/s^2 × 1 J/1 kgm^2/s^2 × 1 kJ/1000 J = 1.59642 kJ

(b) Useful power supplied = work done ÷ time = 1.59642 kJ ÷ 12.3 s = 0.1298 kJ/s × 1 kW/1 kJ/s = 0.1298 kW