A 43 g ice cube can slide without friction up and down a 30∘ slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring 10 cm. The spring constant is 20 N/m . When the ice cube is released, what total distance will it travel up the slope before reversing direction?

51 cm

Explanation:

Givens:  

k = 20 N/m  

x_i,spring = x_is = 10 cm = 0.1 m  

X_f,spring =Xf_s = 0 m

Ф_ramp = B = 30°

m_ice cube = m = 43 g = 0.043 kg

The surface is friction less.  

We will consider the system to be the spring+ the ice cube+the Earth+the friction less surface. There are no external forces that may act on the system. This means that the system is isolated. So that, we can use the conservation of energy principle. Note that ΔE = 0 because there is no energy comes in or out of the system. Note that, we must find the final vertical altitude of the cube, to be able to find the distance traveled up the slope.  

ΔE = 0

Ef —Ei = 0

Kf + Us_f + Ug_f — [ K_i+ U_si + Ug_i] = 0  

Kf + Us_f + Ug_f— K_i - U_si  - Ug_i = 0

1/2 m*v_f^2 +1/2 m*v_f^2 +mgy_f -1/2 m*v_i^2-1/2 k*x_is^2-mgy_i = 0

Note that V_f = 0

find attachment no 2

Note that the spring will return, finally, to its normal length x_fs = 0

find attachment no 3

Note that the initial velocity of the cube is zero too vi = 0 =0  

find attachment no 4

h = 1/2*20*0.1^2/0.043*9.8

h = 0.22

From the geometry of the right triangle, in the figure down:  

sin Ф =h/d

d*sin Ф = h

d = h/sin Ф

d =0.51 m