Question

A 3.20 kg block starts at rest and slides a distance d down a frictionless β = 30.0 ◦ incline, where it runs into a spring with spring constant k = 431 N/m. The block slides an additional 21.0 cm before it is brought to rest momentarily by compressing the spring. (a) What is the value of d? (b) Where is the speed of the block at the maximum value? Is it when the block first comes into contact with the spring? If not, how much further below this point of contact does it travel before it reaches this maximum speed? Compare this point to the point where there is zero net force on the block.

Answer 1

Answer:

Explanation:

a ) work done by gravitational force

= mg sinθ ( d + .21)

Potential energy stored in compressed spring

= 1/2 k x²

= .5 x 431 x ( .21 )²

= 9.5

According to conservation of energy

mg sinθ ( d + .21)  = 9.5

3.2 x 9.8 x sin 30( d + .21 ) = 9.5

d = 40 cm

b )

As long as mg sin30 is greater than kx ( restoring force ) , there will be acceleration in the block.

mg sin30 = kx

3.2 x 9.8 x .5 = 431 x

x =  3.63 cm

When there is compression of 3.63 cm in the spring , block will have maximum velocity. there after its speed will start decreasing.