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?

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Answer 1 · 2017-05-25T14:56:00

$13.7ms^-1$ , rounded to one decimal place.

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