Question

A 2.9 kg block is projected at 5.4 m/s up a plane that is inclined at 40∘ with the horizontal

a
How far up along the plane does the block go if the coefficient of kinetic fraction between the block and the plane is 0.375?

b..How far up the plane does the block go if
If the block then slides back down the plane, what is its speed when it returns to its original projection point?the plane is frictionless? Give your answer with three significant figures.

c.

Answer 1

a)

from the force diagram , force equation perpendicular to incline is given as

F_n = mg Cos40

u_k = Coefficient of kinetic friction = 0.375

kinetic frictional force acting on the block is given as

f_k = u_kF_n = u_k mg Cos40

d = distance moved by the block along the incline before stopping

work done by the frictional force is given as

W_k = f_k d Cos180 = (u_k mg Cos40) (- d) = - u_k mgd Cos40

v_i = initial speed at the bottom of the incline = 5.4 m/s

v_f = final speed at the top of incline = 0 m/s

h = height gained by the block = d Sin40

using conservation of energy between top and bottom of incline

kinetic energy at bottom + work done by frictional force = potential energy at the top + kinetic energy at top

(0.5) m v_i² + W_k = mgh + (0.5) m v_f²

(0.5) m v_i² - u_k mgd Cos40 = mgh + (0.5) m v_f²

(0.5) v_i² - u_k gd Cos40 = gd Sin40 + (0.5) v_f²

(0.5) (5.4)² - (0.375 x 9.8 Cos40) d = (9.8 Sin40) d + (0.5) (0)²

d = 1.60 m

b)

v_f' = final speed at the bottom

v_i' = initial speed at the top = 0 m/s

using conservation of energy between top and bottom of incline

potential energy at the top + work done by frictional force = kinetic energy at the bottom

mgh + W_k = (0.5) m v'_i²

mgd Sin40 - u_k mgd Cos40 = (0.5) m v'_i²

gd Sin40 - u_k gd Cos40 = (0.5) v'_i²

(9.8 x 1.6 Sin40) - (0.375 x 9.8 x 1.6 Cos40) = (0.5) v'_i²

v'_i = 3.34 m/s