Question

The 16.0-in. spring is compressed to a 9.9-in. length, where it is released from rest and accelerates the sliding block A. The acceleration has an initial value of 420 ft/sec2 and then decreases linearly with the x-movement of the block, reaching zero when the spring regains its original 16.0-in. length. Calculate the time t for the block to go (a) 3.05 in. and (b) 6.1 in.

Answer 1

Answer:

Part a : Time required for the block to go to 3.05 in is 0.036 s

Part b : Time required for the block to go to 6.1 in is 0.054 s

Explanation:

The acceleration is given as

$a=a_o-kx$

Here

• $a_o$ is the initial acceleration which is given as 420 $ft/s^2$
• k is the spring constant which is calculated as follows

                                      $k=\frac{a_o}{l_c-l_{uc}}$

• Here lc is the compressed length of the spring
• and Iuc is the uncompressed length of the spring so  the value of k is

                                   $k=\frac{420}{\frac{16-9.9}{12}}\\k=826.22 s^{-2}$

So now the equation becomes

$a=420-826.23x$

As per relation of acceleration to velocity

$vdv=adx$

$\int\limits^v_0 {v} \, dv =\int a dx$

$\frac{v^2}{2} =\int (420-826x) dx$

$\frac{v^2}{2} =420x-413x^2$

${v^2}=840x-826x^2\\{v}=\sqrt{840x-826x^2}$

Also

$v=\frac{dx}{dt}$

So

$\frac{dx}{dt}=\sqrt{840x-826x^2}$

Solving for dt and integrating results in

${dt}=\frac{dx}{\sqrt{840x-826x^2}}\\\int\limits^t_0 \,dt=\int\limits^x_0 \,\frac{dx}{\sqrt{840x-826x^2}}\\t=\int\limits^x_0 \,\frac{dx}{\sqrt{840x-826x^2}}$

Now using this equation value for both parts can be calculated numerically as

Part a 3.05 in

$x=\frac{3.05}{12}\\x=0.254 ft$

Solving numerically for this value of x

$t=\int\limits^{0.254}_0 \,\frac{dx}{\sqrt{840x-826x^2}}\\t=0.036 s$

Time required for the block to go to 3.05 in is 0.036 s

Part b 6.1 in

$x=\frac{6.1}{12}\\x=0.504 ft$

Solving for this value of x

$t=\int\limits^{0.504}_0 \,\frac{dx}{\sqrt{840x-826x^2}}\\t=0.0543 s$

Time required for the block to go to 6.1 in is 0.054 s