A box of mass 180 kg moves down a smooth ramp with an initial speed of 6.00 m/s.

Question

A box of mass 180 kg moves down a smooth ramp with an initial speed of 6.00 m/s. The
ramp is 3.70 m long and is inclined at 39.0 degrees to horizontal. Choose y=0 at the bottom and
using energy methods:
i. Determine the initial gravitational potential energy of the box. [4110 J]
ii. Determine the initial kinetic energy of the box. [3240 J]
iii. Find the speed of the box when it reaches the bottom of the ramp. [9.04 m/s]
iv. Find the distance the box travels on the floor before it stops, if the coefficient of kinetic
friction between the box and the floor is 0.28. [14.9 m]

Explanation / Answer

i)

Gravitational potential energy = mgh = 180*9.8*3.7 sin 39 = 4107 Joules

ii)

Initial KE = 1/2 mv^2 = 0.5*180*6^2 = 3240 J

iii)

We have

v^2 = u^2 +(2ax) = 36+sqrt(2*9.8*3.7 ) = 9.04 m/s

or we can calculate the time taken, t and then, ubtitute in v = u+at

iv)

Kinetic energy = Work done

0.5 m*v^2 = mg*Mu*x

x = 0.5 v^2/Mu*g = 14.89 m ( v = 9.04 m/s)

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