Work and Energy

A force of 50 N acts on the block at the angle shown in the diagram. The block moves a horizontal distance of 3.0 m. Calculate the work being done by the force.

Answer:

Work done,
W = F × s × cos θ
W = 50 × 3.0 × cos30o = 129.9J

Diagram above shows a 10N force is pulling a metal. The friction between the block and the floor is 5N. If the distance travelled by the metal block is 2m, find

1. the work done by the pulling force
2. the work done by the frictional force

Asnwer:

(a) The force is in the same direction of the motion. Work done by the pulling force,

      W = F × s = (10)(2) = 20J

(b) The force is not in the same direction of motion, work done by the frictional force

      W = F × s × cos180o= (5)(2)(-1) = -10J

Answer:

In this case, Ranjit does work to overcome the gravity.
Ranjit's mass = 45kg
Vertical height of the motion, h = 35 × 0.15
Gravitational field strength, g = 10 ms-2
Work done, W = ?

W = mgh = (45)(10)(35 × 0.15) = 2362.5J

The graph above shows the force acting on a trolley of 5 kg mass over a distance of 10 m. Find the work done by the force to move the trolley.

Answer:

In a Force-Displacement graph, work done is equal to the area below the graph. Therefore, work done

            W = \frac{1}{2} \times 8 \times 10 = 40J

Determine the kinetic energy of a 2000-kg bus that is moving with a speed of 35.0 m/s.

Answer:

Kinetic Energy,

            \begin{gathered}
  E_K  = \frac{1}
{2}mv^2  \hfill \\
  E_K  = \frac{1}
{2}(2000)(35)^2  \hfill \\
  E_K  = 1225000J \hfill \\ 
\end{gathered}  

A ball of 1kg mass is droppped from a height of 4m. What is the maximum kinetic energy possessed by the ball before it reached the ground?

Answer:

According to the principle of conservation of energy, the amount of potential energy losses is equal to the amount of kinetic energy gain.

Maximum kinetic energy
= Maximum potentila energy losses
= mgh = (1)(10)(4) = 40J

Diagram above shows a spring with a load of mass 0.5kg. The extention of the spring is 6cm, find the energy stored in the spring.

Answer:

The energy stored in the spring is the elestic potential energy.

            \begin{gathered}
  E_P  = \frac{1}
{2}Fx \hfill \\
  E_P  = \frac{1}
{2}(5)(0.06) = 0.15J \hfill \\ 
\end{gathered} 
            

A trolley of 5 kg mass moving against friction of 5 N. Its velocity at A is 4ms-1 and it stops at B after 4 seconds. What is the work done to overcome the friction?

Answer:

In this case, kinetic energy is converted into heat energy due to the friction. The work done to overcome the friction is equal to the amount of kinetic energy converted into heat energy, hence

              \begin{gathered}
  {\text{Work done}} \hfill \\
  {\text{ =  Kinetic energy losses}} \hfill \\
  {\text{ =  }}\frac{1}
{2}mv_1^2  - \frac{1}
{2}mv_2^2  \hfill \\
   = \frac{1}
{2}(5)(4)^2  - \frac{1}
{2}(5)(0) \hfill \\
   = 40J \hfill \\ 
\end{gathered}