Chapter 11, Problem 63A

(a)

To determine

The work done by hoisting the sack of grain in to the storage room.

Answer to Problem 63A

  $W=4.9\times {10}^3\text{ J}$

Explanation of Solution

Given:

Weight of the sack of grain is $mg=98.0\text{ N = 98}\text{.0 kg}{\text{.m/s}}^2$ .

Height of the storage room above the ground floor of a grain elevator is $\Delta h=50.0\text{ m}$ .

Formula used:

Work done by hoisting the sack of grain in to the storage room can be calculated using the equation,

  $W=mg\Delta h$  $\left(1\right)$

Calculation:

Substituting the numerical values in equation $\left(1\right)$ ,

  $W=\left(\text{98}\text{.0 kg}{\text{.m/s}}^2\right)\left(50.0\text{ m}\right)$

  $=4.9\times {10}^3\text{ kg}{\text{.m}}^2{\text{/s}}^2$

  $=4.9\times {10}^3\text{ J}$

Conclusion:

The work done by hoisting the sack of grain in to the storage room is $4.9\times {10}^3\text{ J}$ .

(b)

To determine

The increase in potential energy of a system containing the sack of grain and Earth.

Answer to Problem 63A

  $\Delta PE=4.9\times {10}^3\text{ J}$

Explanation of Solution

Given:

Weight of the sack of grain is $mg=98.0\text{ N = 98}\text{.0 kg}{\text{.m/s}}^2$

Height of the storage room above the ground floor of a grain elevator is $\Delta h=50.0\text{ m}$

Formula used:

The increase in potential energy of a system containing the sack of grain and Earth can be calculated using the equation,

  $\Delta PE=P{E}_{sr}-P{E}_E$  $\left(2\right)$

Where,

$P{E}_{sr}$ is the potential energy of the sack of grain at the storage room, $P{E}_{sr}=mg\Delta h$

and

$P{E}_E$ is the potential energy of the sack of grain on the Earth, $P{E}_E=0$ since height is zero.

Substituting these values in equation $\left(2\right)$ ,

  $\Delta \text{PE = }mg\Delta h-0=mg\Delta h$  $\left(3\right)$

Calculation:

Substituting the numerical values in equation $\left(3\right)$ ,

  $\Delta PE=\left(\text{98}\text{.0 kg}{\text{.m/s}}^2\right)\left(50.0\text{ m}\right)$

  $=4.9\times {10}^3\text{ kg}{\text{.m}}^2{\text{/s}}^2$

  $=4.9\times {10}^3\text{ J}$

Conclusion:

The increase in potential energy of a system containing the sack of grain and Earth is $4.9\times {10}^3\text{ J}$ .

(c)

To determine

The kinetic energy of the sack of grain just before it strikes the ground floor.

Answer to Problem 63A

  $KE=4.9\times {10}^3\text{ J}$

Explanation of Solution

Given:

Weight of the sack of grain is $mg=98.0\text{ N = 98}\text{.0 kg}{\text{.m/s}}^2$

Height of the storage room above the ground floor of a grain elevator is $\Delta h=50.0\text{ m}$

Formula used:

Kinetic energy of a system is

  $KE=\frac{1}{2}m{v}^2$

Calculation:

When an object is at rest at a height from the ground, it possesses some potential energy but its kinetic energy is zero. If it falls from the height to the ground, it will lose its potential energy and converted into kinetic energy.

Therefore, the kinetic energy of the sack of grain just before striking the ground floor is equal to the potential energy of the sack of grain at height $\Delta h$ from the ground floor.

So, the kinetic energy of the sack of grain just before it strikes the ground floor is,

  $KE=\Delta PE=4.9\times {10}^3\text{ J}$

Conclusion:

The kinetic energy of the sack of grain is $4.9\times {10}^3\text{J}$ .