A stone of mass 20g is projected vertically upward with a catapult whose rubber has a force constant of 72Nm-1. If the tension in the cord at the point of release is 36N, find
(a) the velocity of projection of the stone
(b) the maximum height attained by the stone
Pls show the workings
Plz the answer
Use V=square root of FL/M. You can get your F from F=ma and L=e/k
jack
a cataputt is used to project a stone of mass 50g. if the rubber of the catapult has an elastic constant of 200-1 and was stretched 5cm. what speed can it given to the mass
It always help to draw a diagram
Hooke's law F=kx
x is the extension of the spring
Force F is acting on the stone when it is still in the catapult, travelling distance x.
x=F/k=36/72
Initial velocity u = 0
"Final" velocity v = ?
Acceleration a = F/m - mg
Displacement x
a) We can use kinematics formula v^2=u^2-2as
Sub in the values to get v
b) Use the same formula. This time, the final velocity is 0(At maximum height). Our initial velocity is the velocity found in (a). a is now only the weight - no other force is acting on the stone.