A 1.50-kg block is placed on a platform on top of a vertical spring. With the bl

Question

A 1.50-kg block is placed on a platform on top of a vertical spring. With the block on top, the spring is compressed a distance of x = 10.0 cm and then the block is released from rest from this position (as shown in the image on the left). The block is seen to travel to a maximum height of h = 13.0 cm above the top of the spring platform before stopping, turning around and falling back down. What was the speed of the block as it left the platform? (You can assume the block left the platform at the same height as the top of the pink columns on either side of the spring, as shown in the image on the right.) IMAGE HERE: http://imgur.com/Tlj8aFV

Explanation / Answer

Force in Newtons, N
= Mass kg * Earth's Gravitational Constant g in N/kg
= 1.50*9.8 = 14.7N

10cm = 0.1m

From Hookes Law F = k * x .
F=Force in Newtons, k =Spring Constant N/m, x =Displacement in metres,

k = F/x = 14.7/0.1m = 1.47 N/m

now we conserve the energy we get the equation as

(1/2)kx^2 + mgh = (1/2) mv^2

put hte values we get

(1/2)*1.47*(0.1)^2 + 1.5*9.8*0.13 = (1/2) *1.5*v^2

0.00735 + 1.911 = 0.75 * v^2

1.91835 = 0.75*v^2

2.5578 = v^2

v = 1.59 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.