Question

An 57.0 N box of clothes is pulled 8.1 m up a 30.0° ramp by a force of 106 N that points along the ramp. If the coefficient of kinetic friction between the box and the ramp is 0.22, calculate the change in the box's kinetic energy.

Answer 1

The change in the box's kinetic energy, considering work against friction, work by the applied force, and change in gravitational potential energy, is approximately 2243.14 J.

To calculate the change in the box's kinetic energy, we need to consider the work done by the applied force, the work done against friction, and the change in gravitational potential energy.

The work done by the applied force (parallel to the ramp) is given by:

`\[ W_{\text{applied}} = F_{\text{applied}} \cdot d \cdot \cos(\theta) \]`

where

`\[ F_{\text{applied}} = 106 \, \text{N} \]` (applied force),

`\[ d = 8.1 \, \text{m} \]` (distance), and

`\[ \theta = 30.0^\circ \]`(angle between the force and the direction of motion).

The work done against friction is given by:

`\[ W_{\text{friction}} = -\mu_k \cdot N \cdot d \]`

where

`\[ \mu_k = 0.22 \]` (coefficient of kinetic friction),

N is the normal force (equal to the component of the weight perpendicular to the ramp), and

`\[ N = mg \cdot \cos(\theta) \]` (where m is the mass and g is the acceleration due to gravity).

The change in gravitational potential energy is given by:

`\[ \Delta PE = m \cdot g \cdot \Delta h \]`

where

`\[ \Delta h \]` is the change in height, which is
`\( d \cdot \sin(\theta) \)`.

Now, we can calculate the change in kinetic energy:

`\[ \Delta KE = W_{\text{applied}} + W_{\text{friction}} + \Delta PE \]`

Let's perform the calculations:

`\[ W_{\text{applied}} = 106 \, \text{N} \cdot 8.1 \, \text{m} \cdot \cos(30.0^\circ) \approx 874.36 \, \text{J} \]\[ N = (57.0 \, \text{N}) \cdot 9.8 \, \text{m/s}^2 \cdot \cos(30.0^\circ) \approx 490.24 \, \text{N} \]`

`\[ W_{\text{friction}} = -0.22 \cdot 490.24 \, \text{N} \cdot 8.1 \, \text{m} \approx -869.91 \, \text{J} \]\[ \Delta h = 8.1 \, \text{m} \cdot \sin(30.0^\circ) \approx 4.05 \, \text{m} \]\[ \Delta PE = (57.0 \, \text{N}) \cdot 9.8 \, \text{m/s}^2 \cdot 4.05 \, \text{m} \approx 2238.69 \, \text{J} \]`

Now, calculate the change in kinetic energy:

`\[ \Delta KE = 874.36 \, \text{J} - 869.91 \, \text{J} + 2238.69 \, \text{J} \approx 2243.14 \, \text{J} \]`

Therefore, the change in the box's kinetic energy is approximately
`\(2243.14 \, \text{J}\)`.