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
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 F_g= m g sin theta
where sin theta = 30 / 400 = 0.075

Inserting given vales we get
F_g= 500xx9.81 xx0.075=367.875 N

Net force F_g-F_f= 367.855- 250 = 117.875 N

From Newton's Second Law of motion
a =F/m= 117.875 / 500 = 0.23575 ms^-2

Using the following kinematic expression
v^2-u^2=2as
Given initial velocity of truck is = 0, and inserting various values we get

v_f^2 - 0^2 = 2 a s
=>v_f^2 = 2 xx0.23575 xx400
=>v_f = sqrt(188.6)=13.7ms^-1, rounded to one decimal place.