A truck of mass 500kg moves from rest at the top of a section track 400m long and 30m high as shown. The frictional force acting on the truck is 250N throughout this journey. What is the final speed of the truck?

1 Answer
A08
May 25, 2017

13.7ms−1, rounded to one decimal place.

Explanation:

Two forces are acting on the truck at any point of time during its motion.

  1. weight =mg, acting downwards.
  2. Force due to friction. Acting along the track and opposing the motion.

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Therefore, net force on the truck is the difference of the component of its weight parallel to the track and the frictional force.

Component of gravitational force parallel to track Fg=mgsinθ
where sinθ=30400=0.075

Inserting given vales we get
Fg=500×9.81×0.075=367.875N

Net force Fg−Ff=367.855−250=117.875N

From Newton's Second Law of motion
a=Fm=117.875500=0.23575ms−2

Using the following kinematic expression
v2−u2=2as
Given initial velocity of truck is =0, and inserting various values we get

v2f−02=2as
⇒v2f=2×0.23575×400
⇒vf=√188.6=13.7ms−1, rounded to one decimal place.

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