(a) For the work-energy theorem, the work done to lift the can of paint is equal to the gravitational potential energy gained by it, therefore it is equal to

where m=3.4 kg is the mass of the can, g=9.81 m/s^2 is the gravitational acceleration and
is the variation of height. Substituting the numbers into the formula, we find

(b) In this case, the work done is zero. In fact, we know from its definition that the work done on an object is equal to the product between the force applied F and the displacement:

However, in this case there is no displacement, so d=0 and W=0, therefore the work done to hold the can stationary is zero.
(c) In this case, the work done is negative, because the work to lower the can back to the ground is done by the force of gravity, which pushes downward. Its value is given by the same formula used in part (a):

Answer:
1176 Nm or J
Explanation:
W = F*d
F = 60kg * 9.8 kgm/s^2 = 588 N
W = 588 N * 2m = 1176 N*m
Answer:
the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is 
Explanation:
The free-body diagram below shows the interpretation of the question; from the diagram , the wheel that is rolling in a clockwise directio will have two velocities at point P;
- the peripheral velocity that is directed downward
along the y-axis
- the linear velocity
that is directed along the x-axis
Now;


Also,

where
(angular velocity) = 

∴ the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is 