Moving a block up a Plane Question?

Question

A block is given an initial speed of 3.0 m/s up the 22 degree plane. How far up the plane will it go? How much time elapses befoer it returns to starting point. Asumme Uk is 0.17

Answer 1

Nikhil007 forgot about any energy coverting to potential.

The initial kinetic energy will be converted to (1) work against friction and (2) potential energy.

(1/2)*m*v^2 = uk*N*d + m*g*y
where N is the normal force, d is the distance along the plane, and y is vertical distance.

N = m*g*cos22
y = h*sin22

So, going back to
(1/2)*m*v^2 = uk*N*d + m*g*y
(1/2)*m*(3 m/s)^2 = 0.17*m*g*cos22*d + m*g*d*sin22
(0.17*cos22 + sin22)g*d = 4.5 m^2/s^2
d = (4.5 m^2/s^2) / (.53*9.8 m/s^2) = .86 m

2nd part:
The component of weight down the slope:
m*g*sin22 = 3.67 (m/s^2)*m
The force of friction
0.17*m*g*cos22 = 1.67 (m/s^2)*m
Net force down the slope = 2 (m/s^2)*m
Acceleration down the slope
F = 2 (m/s^2)*m = m*a
a = 2 m/s^2

Time to go from a stop to the starting point (.86 m down again)
d = (1/2)*a*t^2
.86 m = (1/2)*2 m/s^2*t^2
t^2 = .86 s^2 = 0.93 s