We can use the principle of conservation of energy to find the velocity of the stone when released.
The potential energy stored in the rubber band when it is stretched is given by:
PE = 1/2 k x^2
where k is the spring constant of the rubber band and x is the distance it is stretched.
Assuming the rubber band behaves like a spring, we can use Hooke's law to calculate the spring constant:
F = kx
where F is the force applied and x is the displacement.
Substituting the given values, we get:
k = F/x = 70/0.07 = 1000 N/m
Now we can calculate the potential energy of the rubber band:
PE = 1/2 k x^2 = 1/2 * 1000 * (0.07)^2 = 2.45 J
When the stone is released, all the potential energy is converted to kinetic energy:
KE = 1/2 mv^2
where m is the mass of the stone and v is its velocity.
Equating PE and KE, we get:
1/2 mv^2 = 2.45
v^2 = 4.9/0.005 = 980
v = √980 = 31.3 m/s
Therefore, the instantaneous velocity of the stone when released is 31.3 m/s.
A stone of mass 5g is projected with a rubber catapult. If the catapult is stretched through a distance of 7cm by a force of 70N .
calculate the instantaneous velocity of the stone when released
1 answer
1 answer