0 = 30°. The box has a mass m = 4 kg, the coefficient of kinetic friction between the box the inclined is µ = 0.5, and the block moves a distance d = 10 m up the inclined plane. What is the work done by the force F? Answer: 400 J. a) b) What is the normal force from the inclined plane? Answer: 33.9 N c) What is the work done by the kinetic friction force? Answer: - 169.5 J d What is bu 106 1

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In the figure below, we have a force \( F = 40 \, \text{N} \) pushing a box up an inclined plane with an angle \( \theta = 30^\circ \). The box has a mass \( m = 4 \, \text{kg} \), the coefficient of kinetic friction between the box and the inclined plane is \( \mu_k = 0.5 \), and the block moves a distance \( d = 10 \, \text{m} \) up the inclined plane.

a) What is the work done by the force \( F \)?  
**Answer:** 400 J.

b) What is the normal force from the inclined plane?  
**Answer:** 33.9 N.

c) What is the work done by the kinetic friction force?  
**Answer:** -169.5 J.

d) What is the work done by the gravitational force?  
**Answer:** -196 J.

e) What is \( W_{\text{net}} \), the net work done on the box?  
**Answer:** 34.5 J.

f) The box starts from rest. What is the speed of the box after it has travelled the distance \( d = 10 \, \text{m} \) up the incline plane?  
**Answer:** 4.2 m/sec.

The diagram shows a box on an inclined plane angled at \( 30^\circ \) with a force of 40 N applied parallel to the incline. The box's mass is 4 kg, and the kinetic friction coefficient is 0.5.
Transcribed Image Text:In the figure below, we have a force \( F = 40 \, \text{N} \) pushing a box up an inclined plane with an angle \( \theta = 30^\circ \). The box has a mass \( m = 4 \, \text{kg} \), the coefficient of kinetic friction between the box and the inclined plane is \( \mu_k = 0.5 \), and the block moves a distance \( d = 10 \, \text{m} \) up the inclined plane. a) What is the work done by the force \( F \)? **Answer:** 400 J. b) What is the normal force from the inclined plane? **Answer:** 33.9 N. c) What is the work done by the kinetic friction force? **Answer:** -169.5 J. d) What is the work done by the gravitational force? **Answer:** -196 J. e) What is \( W_{\text{net}} \), the net work done on the box? **Answer:** 34.5 J. f) The box starts from rest. What is the speed of the box after it has travelled the distance \( d = 10 \, \text{m} \) up the incline plane? **Answer:** 4.2 m/sec. The diagram shows a box on an inclined plane angled at \( 30^\circ \) with a force of 40 N applied parallel to the incline. The box's mass is 4 kg, and the kinetic friction coefficient is 0.5.
The image contains equations related to work in physics:

\[ W = F_{\parallel} \Delta x = \vec{F} \cdot \Delta \vec{x} \]

This equation represents the work done \( W \) by a force \( F \) that is parallel to the displacement \( \Delta x \). It can also be expressed as the dot product of the force vector \( \vec{F} \) and the displacement vector \( \Delta \vec{x} \).

\[ W_{\text{net}} = \frac{1}{2} m (v_f)^2 - \frac{1}{2} m (v_i)^2 \]

This relates the net work \( W_{\text{net}} \) to the change in kinetic energy of an object, where \( m \) is the mass, \( v_f \) is the final velocity, and \( v_i \) is the initial velocity.

\[ W = \int F_x \, dx \]

This integral represents the work done when a varying force \( F_x \) acts along the x-axis over a distance.

\[ W = \int \vec{F} \cdot d\vec{r} \]

This integral represents the work done by a force \( \vec{F} \) along a path, where \( d\vec{r} \) is an infinitesimal displacement vector.

These equations explore the concept of work done by forces on an object and relate it to displacement and energy changes.
Transcribed Image Text:The image contains equations related to work in physics: \[ W = F_{\parallel} \Delta x = \vec{F} \cdot \Delta \vec{x} \] This equation represents the work done \( W \) by a force \( F \) that is parallel to the displacement \( \Delta x \). It can also be expressed as the dot product of the force vector \( \vec{F} \) and the displacement vector \( \Delta \vec{x} \). \[ W_{\text{net}} = \frac{1}{2} m (v_f)^2 - \frac{1}{2} m (v_i)^2 \] This relates the net work \( W_{\text{net}} \) to the change in kinetic energy of an object, where \( m \) is the mass, \( v_f \) is the final velocity, and \( v_i \) is the initial velocity. \[ W = \int F_x \, dx \] This integral represents the work done when a varying force \( F_x \) acts along the x-axis over a distance. \[ W = \int \vec{F} \cdot d\vec{r} \] This integral represents the work done by a force \( \vec{F} \) along a path, where \( d\vec{r} \) is an infinitesimal displacement vector. These equations explore the concept of work done by forces on an object and relate it to displacement and energy changes.
Expert Solution
Step 1: Given Data:

Given 

The mass of the box is 

                    m space equals space 4 space k g

Applied force is

                     F equals 40 space N

The angle of inclination is

                      theta equals 30 degree

The coefficient of kinetic friction between the box and the inclined plane is

                      mu subscript k equals 0.5 space

Displacement of the box is

                     d equals 10 space m



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