A 7.3 kg blocks sits on a flat, frictionless surface and has been pushed against

Question

A 7.3 kg blocks sits on a flat, frictionless surface and has been pushed against a spring. The spring (k = 220 N/m) has been compressed by 3.5 cm from its equilibrium position by the block. There is a 12 degree, rough incline that starts 10 m away from the spring, as shown in the figure below. Note that µk = 0.58 on the incline.

a) Determine the speed of the block after it leaves the spring.

b) How far up the incline will the block travel?

c) Determine if the block will reach the bottom of the incline after reaching its maximum height and, if so, what speed will the block have as it leaves the incline on the way back down.

Explanation / Answer

Energy stored in Compressed string = 0.5*k*x^2
where k = 220 N/m
x = 3.5 * 10^-2 m
Kinetic Energy gained by the block = 0.5*m*v^2
where m = 7.3 kg
v = Speed of the block as it leaves spring.

0.5*m*v^2 =  0.5*k*x^2
v = x * sqrt(k/m)
v = 3.5 * 10^-2 * sqrt(220/7.3) m/s
v = 0.192 m/s
Speed of the block after it leaves the spring, v = 0.192 m/s

(b)
Velocity of block at the start of incline = 0.192 m/s
Final Velocity = 0
Energy lost in Friction = uk * mg*cos(12) * s
Using Energy Conservation,
K.Ein + P.Ein = Energy lost in Friction + K.Efi + P.Efi
0.5*7.3*0.192^2 + 0 = uk * mg*cos(12) * d + 0 + m*g*h
0.5*7.3*0.192^2 + 0 = 0.58 * 7.3*9.8*cos(12) * d + 0 + 7.3*9.8*h

sin(12) = h/d
h = d*sin(12)

0.5*7.3*0.192^2 + 0 = 0.58 * 7.3*9.8*cos(12) * d + o + 7.3*9.8*d*sin(12)
d = 2.43 * 10^-3 m

Distance moved up the incline, d = 0.243 cm

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