The small (2.3kg) block is placed against a 895 N/m spring that has been compres
ID: 1288832 • Letter: T
Question
The small (2.3kg) block is placed against a 895 N/m spring that has been compressed by 12cm. After the spring is released, the small block slides along the frictionless track until it cloodies with the large (5.1kg) block. The two stick teogerther and slide to the right. To the right of the large block the horizontal surface has a coeff. of friction of 0.23 with the blocks. After the collision, how far do the blocks slide before stopping?
The small (2.3kg) block is placed against a 895 N/m spring that has been compressed by 12cm. After the spring is released, the small block slides along the frictionless track until it cloodies with the large (5.1kg) block. The two stick teogerther and slide to the right. To the right of the large block the horizontal surface has a coeff. of friction of 0.23 with the blocks. After the collision, how far do the blocks slide before stopping?Explanation / Answer
v is the speed og the 2.3 kg after spring is released
0.5*K*X^2 = 0.5*m*v^2
895*.12*.12 = 2.3*v^2
v = 2.37 m/s
after collision
mv = = (M+m)*V
V = mv/(M+m) = 0.74 m/s
using work energy theorem
K1 + W = K2
0.5*(M+m)*V^2 -u*(M+m)*g*d = 0
d = V^2/2ug = (0.74*0.74)/(2*0.23*9.81)
d = 0.121m = 12.1 cm
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