A light sparing with force constant 3.80 N?M IS COMPRESSED by 7.40 am as it is b

Question

A light sparing with force constant 3.80 N?M IS COMPRESSED by 7.40 am as it is between a 0.350-kg block on the left and a 0.700-kg block on the right, both resting on a horizontal surface. The spring exerts a force on each block, tending to push the blocks apart. the blocks are simultaneously released from rest. Find the acceleration with which each positive to the right and negative to the left, Indicate the direction with the sign of answer. Assume that the coefficient of static friction is the same as the coefficient of kinetic function.

Explanation / Answer

k = spring constant = 3.80 N/m

x = compression = 7.4 cm = 0.074 m

a)

F = spring force = k x = 3.80 x 0.074 = 0.2812 N

for lighter block of mass m1 = 0.350 kg

f1 = frictional force = 0

so force equation is given as

F = m1a1

0.2812 = 0.350 a1

a1 = 0.803 m/s2

for lighter block of mass m2 = 0.700 kg

f2 = frictional force = 0

so force equation is given as

F = m2a2

0.2812 = 0.700 a2

a2 = 0.402 m/s2

b)

F = spring force = k x = 3.80 x 0.074 = 0.2812 N

for lighter block of mass m1 = 0.350 kg

f1 = frictional force = u m1 g = 0.056 x 0.35 x 9.8 = 0.192 N

so force equation is given as

F - f1 = m1a1

0.2812 - 0.192 = 0.350 a1

a1 = 0.255 m/s2

for lighter block of mass m2 = 0.700 kg

f2 = frictional force = u m2 g = 0.056 x 0.70 x 9.8 = 0.384 N

a2 = 0  

since frictional force is greater than the spring force

c)

b)

F = spring force = k x = 3.80 x 0.074 = 0.2812 N

for lighter block of mass m1 = 0.350 kg

f1 = frictional force = u m1 g = 0.418 x 0.35 x 9.8 = 1.43 N

a1 = 0 m/s2

since frictional force is greater than the spring force

for lighter block of mass m2 = 0.700 kg

f2 = frictional force = u m2 g = 0.418 x 0.70 x 9.8 = 2.86 N

a2 = 0  

since frictional force is greater than the spring force

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