Question

A 3.8 kg block is projected at 4.4 m/s up a plane that is inclined at 37o with the horizontal. How far up along the plane does the block go:___________
(a) if the plane is frictionless and
(b) if the coefficient of kinetic friction between the block and the plane is 0.47
(c) what is thermal energy generated by friction?

Answer 1

Answer:

Explanation:

a) Deceleration due to gravity along the inclined plane is given by:

  g*Sin(∅)= 9.8*Sin(37)=5.9 m/s^2.

Since acceleration is constant, we can use equation of motion along the plane. $V^{2} = U^{2} +2as$

where V: final velocity, U:initial Velocity, a:Acceleration and s:distance

0=$4.4^{2} + 2*-5.9*s$ (-ve because it is against gravity)

It will go 1.64 meters up the plane

b) Force due to friction = Coeff. of friction * Normal Force

Normal Force = mg*Cos(∅) = 9.8*3.8*Cos(37) = 29.75;

Coeff of friction = 0.47;

Force due to friction = 0.47*29.75 = 13.98 N

Force due to gravity : 3.8*5.9 = 22.42;

Total force after adding friction = 22.42 + 13.98 = 36.4 N

New Acceleration = 36.4/3.8 = 9.58;

Applying the equation of motion used above with acceleration = 9.58;

It will go 1.01 meters up the plane.

c) Thermal energy due to friction = Max potential energy without friction - max potential energy after friction.

Potential energy w.o friction(m*g*h) = 3.9 * 9.8 * 1.64*Sin(37) = 37.72 Joules

Potential energy with friction(m*g*h)= 3.9 * 9.8 * 1.01*Sin(37) = 23.23 Joules

Energy due to friction = 37.72 - 23.23 = 14.49 Joules