Question(II) A grocery cart with mass of 16 kg is pushed at constant speed up a $12^{\circ}$ ramp by a force $F_{\mathrm{P}}$ which acts at an angle of $17^{\circ}$ below the horizontal. Find the work done by each of the forces $\left(m \overrightarrow{\mathbf{g}}, \overrightarrow{\mathbf{F}}_{\mathrm{N}}, \overrightarrow{\mathbf{F}}_{\mathrm{P}}\right)$ on the cart if the ramp is 7.5 m long.

Studdy Solution

STEP 1

1. The mass of the grocery cart is 16 kg.
2. The ramp is inclined at an angle of $12^{\circ}$ .
3. The force $F_{\mathrm{P}}$ is applied at an angle of $17^{\circ}$ below the horizontal.
4. The ramp is 7.5 m long.
5. The cart is moving at a constant speed, implying that the net work done on the cart is zero.
6. We are to find the work done by the gravitational force $m \overrightarrow{\mathbf{g}}$ , the normal force $\overrightarrow{\mathbf{F}}_{\mathrm{N}}$ , and the applied force $\overrightarrow{\mathbf{F}}_{\mathrm{P}}$ .

STEP 2

1. Calculate the work done by the gravitational force.
2. Calculate the work done by the normal force.
3. Calculate the work done by the applied force $F_{\mathrm{P}}$ .

STEP 3

Calculate the work done by the gravitational force.
The gravitational force acts vertically downward. The component of the gravitational force along the ramp is $mg \sin(12^{\circ})$ .
The work done by the gravitational force is given by:
$W_{\mathrm{gravity}} = m g \sin(12^{\circ}) \cdot d$
where $m = 16 \, \text{kg}$ , $g = 9.8 \, \text{m/s}^2$ , and $d = 7.5 \, \text{m}$ .
$W_{\mathrm{gravity}} = 16 \times 9.8 \times \sin(12^{\circ}) \times 7.5$
Calculate the above expression to find $W_{\mathrm{gravity}}$ .

STEP 4

Calculate the work done by the normal force.
The normal force acts perpendicular to the ramp. Since there is no displacement in the direction of the normal force, the work done by the normal force is:
$W_{\mathrm{normal}} = 0$

STEP 5

Calculate the work done by the applied force $F_{\mathrm{P}}$ .
The component of the applied force along the ramp is $F_{\mathrm{P}} \cos(17^{\circ} + 12^{\circ})$ .
Since the cart moves at constant speed, the net work done is zero, so the work done by the applied force must balance the work done by gravity:
$W_{\mathrm{applied}} = -W_{\mathrm{gravity}}$
Thus, the work done by the applied force is equal in magnitude but opposite in sign to the work done by gravity.
The work done by each of the forces is: - Gravitational force: $W_{\mathrm{gravity}}$ - Normal force: $W_{\mathrm{normal}} = 0$ - Applied force: $W_{\mathrm{applied}} = -W_{\mathrm{gravity}}$

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Parents Influencer program Contact Policy Terms