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A boy in a pedal car is moving at a speed of 1.40 m/s at the start of a 2.00 m high and 12.4 m long incline. The total mass is 45.5 kg, air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N, and the speed at the lower end of the incline is 6.90 m/s. Determine the work done (in J) by the boy as the car travels down the incline.

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akande212

Answer:

W = 655J

Explanation:

Workdone = change in kinetic energy + change in potential energy - workdone against friction

= 1/2m(v2²–v1²) + mg(h2 – h1) + Ff×d

Given

m = 45.5kg

v1 = 1.40m/s

v2 = 6.90m/s

h1 = 2.00m

h2 = 0m (bottom of the incline)

Ff = 41N frictional force

d = 12.4m

Workdone = 1/2×45.5×(6.90²–1.40²) + mg(0 – 2.00) + 41×12.4

Workdone = 655J

Ff ×d = work done against friction.

The work done (in J) by the boy as the car travels down the incline is  655J

Calculation of the work done:

Since A boy in a pedal car is moving at a speed of 1.40 m/s at the start of a 2.00 m high and 12.4 m long incline. The total mass is 45.5 kg and the constant friction force of 41.0 N, and the speed at the lower end of the incline is 6.90 m/s.

Also,

Work done = change in kinetic energy + change in potential energy - work done against friction

= 1/2m(v2²–v1²) + mg(h2 – h1) + Ff×d

Here

  • m = 45.5kg
  • v1 = 1.40m/s
  • v2 = 6.90m/s
  • h1 = 2.00m
  • h2 = 0m that should be bottom of the incline
  • Ff = 41N i.e. frictional force
  • d = 12.4m

So,

Workdone = 1/2×45.5×(6.90²–1.40²) + mg(0 – 2.00) + 41×12.4

= 655J

Hence, The work done (in J) by the boy as the car travels down the incline is  655J

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