Question

A block of mass 23 kg is set on top of a 37 degree incline with a hight of 13.5 m. The incline rests on a table that is 27.8 m above the ground on the planet Mercury (M = 3.285 * 10^23 kg, R = 1516 mi). The coefficient of kinetic friction between the block and the incline is 0.18. The gravitational acceleration on Mercury is 3.68 m/s.

How far from the base of the table does the block land?

Answer 1

consider the motion on the incline

h = height of incline = 13.5 m

= angle of incline = 37 deg

L = length of he incline

length of he incline is given as

Sin37 = h/L

L = h/Sin37 = 13.5/Sin37 = 22.4 m

perpendicular to incline , force equation is given as

F_n = mg Cos37 eq-1

u_k = Coefficient of kinetic friction between block and incline = 0.18

kinetic frictional force is given as

f_k = u_k F_n  

using eq-1

f_k = u_k (mg Cos37  )

f_k = 0.18 (23 x 3.68) Cos37

f_k = 12.2 N

consider the motion between Top and bottom of incline

h = height of incline = 13.5 m

L = length of he incline = 22.4 m

v = speed of the block at the bottom of incline

using consservation of energy between top and bottom

Potential energy at the top = kinetic energy at the bottom + work done by friction

mgh = (0.5) m v² + f_k L

23 x 3.68 x 13.5 = (0.5) (23) v² + (12.2 x 22.4)

v = 8.7 m/s

consider the motion of the block after leaving the incline surface :

consider the motion of block in vertical direction

Y_o = initial position of the block at top of table = 27.8 m

Y_f = final position of block = 0 m

V_oy = initial velocity = - v Sin37 = - (8.7) Sin37 = - 5.24 m/s

t = time taken to land

a_y = acceleration = - 3.68 m/s²

Using the kinematics equation

Y = Y_o + V_oy t + (0.5) a_y t²

0 = 27.8 + (- 5.24) t + (0.5) (- 3.68) t²

t = 2.72 sec

consider the motion along the horizontal direction

X_o = initial position of the block at top of table = 0 m

X_f = final position of block = distance from the base of table = ?

V_ox = initial velocity = v Cos37 = (8.73) Cos37 =6.97 m/s

t = time taken to land = 2.72 sec

a_x = acceleration = 0 m/s²

Using the kinematics equation

X = X_o + V_ox t + (0.5) a_x t²

X = 0 + (6.97) (2.72) + (0.5) (0) (2.71)²

X = 18.96 m