Energy conservation tells you that
M*g*H + M*Vo^2/2 = (mu)*M*g*X
Mass M cancels out. Solve for slide distance X.
mu = 0.277
g = 9.81 m/s^2
Vo = 1.87 m/s
A boy shoves his stuffed toy zebra down a friction-less chute, starting at a height of 1.07 m above the bottom of the chute and with an initial speed of 1.87 m/s. The toy animal emerges horizontally from the bottom of the chute and continues sliding along a horizontal surface with coefficient of kinetic friction 0.277. How far from the bottom of the chute does the toy zebra come to rest? Take g = 9.81 m/s2.
Hint: How does the work performed on the toy by the force of friction relate to the requested distance? How do the initial height and speed relate to the frictional work?
3 answers
Thank you so much!!! this really helped!!!
H = 1.07;
k = 0.277
g = 9.81 m/s^2
Vo = 1.87 m/s
we know, M*g*H + M*Vo^2/2 = k*M*g*X
So, x = (g*H + (Vo)^2/2) / (mu * g)
k = 0.277
g = 9.81 m/s^2
Vo = 1.87 m/s
we know, M*g*H + M*Vo^2/2 = k*M*g*X
So, x = (g*H + (Vo)^2/2) / (mu * g)