Question

A 128.0-N carton is pulled up a frictionless baggage ramp inclined at 30.0∘above the horizontal by a rope exerting a 72.0-N pull parallel to the ramp’s surface. If the carton travels 5.20 m along the surface of the ramp, calculate the work done on it by (a) the rope, (b) gravity, (c) the normal force of the ramp. (d) What is the net work done on the carton? (e) Suppose that the rope is angled at 50.0∘above the horizontal, instead of being parallel to the ramp’s surface. How much work does the rope do on the carton in this case?

Answer 1

Answer:

(A) 374.4 J

(B) -332.8 J

(C) 0 J

(D) 41.6 J

(E)  351.8 J

Explanation:

weight of carton (w) = 128 N

angle of inclination (θ) = 30 degrees

force (f) = 72 N

distance (s) = 5.2 m

(A) calculate the work done by the rope

• work done = force x distance x cos θ

• since the rope is parallel to the ramp the angle between the rope and

        the ramp θ will be 0

       work done = 72 x 5.2 x cos 0

       work done by the rope = 374.4 J

(B) calculate the work done by gravity

• the work done by gravity = weight of carton x distance x cos θ

• The weight of the carton = force exerted by the mass of the carton = m x g

• the angle between the force exerted by the weight of the carton and the ramp is 120 degrees.

      work done by gravity = 128 x 5.2  x cos 120

      work done by gravity = -332.8 J

(C) find the work done by the normal force acting on the ramp

• work done by the normal force = force x distance x cos θ

• the angle between the normal force and the ramp is 90 degrees

       

         work done by the normal force = Fn x distance x cos θ

         work done by the normal force = Fn x 5.2 x cos 90

         work done by the normal force = Fn x 5.2 x 0

         work done by the normal force = 0 J

(D)  what is the net work done ?

• The net work done is the addition of the work done by the rope,       gravitational force and the normal force

     net work done = 374.4 - 332.8 + 0 =  41.6 J  

(E) what is the work done by the rope when it is inclined at 50 degrees to the horizontal

• work done by the rope= force x distance x cos θ
• the angle of inclination will be 50 - 30 = 20 degrees, this is because the ramp is inclined at 30 degrees to the horizontal and the rope is inclined at 50 degrees to the horizontal and it is the angle of inclination of the rope with respect to the ramp we require to get the work done by the rope in pulling the carton on the ramp

work done = 72 x 5.2 x cos 20

work done = 351.8 J

Answer 2

(a) The work done by the rope parallel to the ramp is 374.4 J.

(b) The work done by gravity is -332.8 J.

(c) The work done by normal force acting on the ramp is 0.

(d) The net work done on the carton is 41.6 J.

(e) The work done by the rope when it is inclined at 50 degrees to the horizontal is 351.82 J.

Work done by the rope

The work done by the rope parallel to the ramp is calculated as;

W = Fdcosθ = mgcosθ

W = 72 x 5.2 x cos0

W =  374.4 J

Work done by gravity

The work done by gravity is calculated as follows;

W = mgdcosθ

Where;

• θ is the angle between the carton and the ramp = 120⁰

W = 128 x 5.2 cos120

W = -332.8 J

Work done by normal force acting on the ramp

W = Fₙdcosθ

W = 128 x 5.2 x cos90

W = 0  

Net work on the carton

The net work done on the carton is calculated as follows;

net work done = 374.4 - 332.8 + 0 =  41.6 J  

Work done by the rope when it is inclined at 50 degrees to the horizontal

W = Fₙdcosθ

where;

• θ is the angle between the rope and the ramp = 50 - 30 = 20⁰

W = 72 x 5.2 x cos(20)

W = 351.82 J

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