A car of mass 700 kg moves up a rough inclined plane as shown in the diagram below. What is the net work done…

Question

A car of mass 700 kg moves up a rough inclined plane as shown in the diagram below. What is the net work done on the car if the car moves up the inclined plane at CONSTANT velocity?

ScanSolve AI Tutor · Accepted Answer

here are the solutions to your physics questions:

5.1

If the car moves up the inclined plane at CONSTANT velocity, its kinetic energy does not change. According to the Work-Energy Theorem, the net work done on an object is equal to the change in its kinetic energy. Therefore, the net work done on the car is zero.

5.2

Weight ( $F_g$ or $mg$ ): Acts vertically downwards from the center of mass.
Normal Force ( $F_N$ or $N$ ): Acts perpendicular to the inclined surface, upwards from the surface.
Engine Force ( $F_{engine}$ ): Acts parallel to the inclined surface, upwards along the incline (in the direction of motion).
Kinetic Friction Force ( $f_k$ ): Acts parallel to the inclined surface, downwards along the incline (opposing motion).

graph TD
    A[Car] -->|F_engine (up incline)| B(Motion)
    A -->|F_N (perpendicular to incline)| C(Surface)
    A -->|mg (vertically down)| D(Earth)
    A -->|f_k (down incline)| E(Surface)

    style A fill:#fff,stroke:#333,stroke-width:2px
    style B fill:#fff,stroke:#333,stroke-width:0px
    style C fill:#fff,stroke:#333,stroke-width:0px
    style D fill:#fff,stroke:#333,stroke-width:0px
    style E fill:#fff,stroke:#333,stroke-width:0px

    subgraph Free Body Diagram
        direction TD
        Car_CM(Center of Mass)
        Car_CM -->|$$F_{engine}$$| Up_Incline
        Car_CM -->|$$F_N$$| Perpendicular_Up
        Car_CM -->|$$mg$$| Vertically_Down
        Car_CM -->|$$f_k$$| Down_Incline
    end

(Note: The mermaid diagram is a textual representation. In a hand-drawn diagram, these would be arrows originating from the center of mass of the car.)

5.3.1

The Work-Energy Theorem states that the net work done on an object is equal to the change in its kinetic energy.

5.3.2

Calculate the normal force ( $N$ ).
The normal force is perpendicular to the incline and balances the perpendicular component of gravity.
$N = mg \cos \theta$
Given: $m = 700 \text{ kg}$ , $g = 9.8 \text{ m/s}^2$ , $\theta = 30^\circ$ .
$N = (700 \text{ kg})(9.8 \text{ m/s}^2) \cos 30^\circ$ $N = 6860 \text{ N} \times 0.8660$ $N = 5940.76 \text{ N}$
Calculate the kinetic friction force ( $f_k$ ).
$f_k = \mu_k N$
Given: $\mu_k = 0.32$ .
$f_k = 0.32 \times 5940.76 \text{ N}$ $f_k = 1901.04 \text{ N}$
Calculate the work done by each force.
The car moves a distance $d = 70 \text{ m}$ up the incline.
Work done by the engine ( $W_{engine}$ ):
$W_{engine} = F_{engine} \cdot d$
Given: $F_{engine} = 6000 \text{ N}$ .
$W_{engine} = (6000 \text{ N})(70 \text{ m})$ $W_{engine} = 420000 \text{ J}$
Work done by gravity ( $W_g$ ):
Gravity acts downwards, opposing the upward motion. The component of gravity parallel to the incline is $mg \sin \theta$ .
$W_g = -mg \sin \theta \cdot d$ $W_g = -(700 \text{ kg})(9.8 \text{ m/s}^2) \sin 30^\circ (70 \text{ m})$ $W_g = -6860 \text{ N} \times 0.5 \times 70 \text{ m}$ $W_g = -240100 \text{ J}$
Work done by friction ( $W_f$ ):
Friction opposes motion, so it acts down the incline.
$W_f = -f_k \cdot d$ $W_f = -(1901.04 \text{ N})(70 \text{ m})$ $W_f = -133072.8 \text{ J}$
Calculate the net work done ( $W_{net}$ ).
$W_{net} = W_{engine} + W_g + W_f$ $W_{net} = 420000 \text{ J} - 240100 \text{ J} - 133072.8 \text{ J}$ $W_{net} = 46827.2 \text{ J}$
Apply the Work-Energy Theorem to find the final velocity ( $v_f$ ).
$W_{net} = \Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$
The car starts from rest, so $v_i = 0$ .
$W_{net} = \frac{1}{2}mv_f^2$ $46827.2 \text{ J} = \frac{1}{2}(700 \text{ kg})v_f^2$ $46827.2 \text{ J} = 350 \text{ kg} \cdot v_f^2$ $v_f^2 = \frac{46827.2 \text{ J}}{350 \text{ kg}}$ $v_f^2 = 133.792$ $v_f = \sqrt{133.792}$ $v_f = \boxed{\text{11.57 m/s}}$

A car of mass 700 kg moves up a rough inclined plane as shown in the diagram below. What is the net work done on the car if the car moves up the inclined plane at CONSTANT velocity?

Answer

Step-by-step solution

5.1

5.2

5.3.1

5.3.2

Calculate the normal force ( $N$ ).

Calculate the kinetic friction force ( $f_k$ ).

Calculate the work done by each force.

Calculate the net work done ( $W_{net}$ ).

Apply the Work-Energy Theorem to find the final velocity ( $v_f$ ).

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More Physics Questions

A car of mass 700 kg moves up a rough inclined plane as shown in the diagram below. What is the net work done on the car if the car moves up the inclined plane at CONSTANT velocity?

Answer

Step-by-step solution

5.1

5.2

5.3.1

5.3.2

Calculate the normal force (NNN).

Calculate the kinetic friction force (fkf_kfk​).

Calculate the work done by each force.

Calculate the net work done (WnetW_{net}Wnet​).

Apply the Work-Energy Theorem to find the final velocity (vfv_fvf​).

Try one yourself

Still stuck on this one?

More Physics Questions

Calculate the normal force ( $N$ ).

Calculate the kinetic friction force ( $f_k$ ).

Calculate the net work done ( $W_{net}$ ).

Apply the Work-Energy Theorem to find the final velocity ( $v_f$ ).