Question

A force F is used to raise a 4-kg mass M from the ground to
a height of 5 m.
What is the work done by the force F? (Note : sin 60° = 0.87;
cos 60° = 0.50. Ignore friction and the weights of the pulleys)
(a) 50 J (b) 100 J
(c) 174 J (d) 200 J

Answer 1

Answer:

2s=2×5=10

w=F×S=4×10×5

Answer 2

The work done by the force is 200J.

Given:-

Mass = 4kg

Height to which it is raised = 5m

Sin 60° = 0.87

Cos 60° = 0.50

To Find:-

The work done by the force.

Solution:-

We can easily calculate the value of work done by the force by using these simple steps.

As

Mass = 4kg

Height to which it is raised = 5m

Sin 60° = 0.87

Cos 60° = 0.50

g = 10m/s²

Now, According to the formula,

Work done = change in potential energy

$W = ∆U$

$W =mgh$

on putting the values, we get

$W = 4 \times 10 \times 5$

$W = 40 \times 5$

$W = 200j$

Hence, The work done by the force is 200J.

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