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.
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
7 answers
A catapult is use to project a stone of a mass 50kg.If the rubber of the catapult has an elastic constant of 200Nm ^-1and was stretch 5cm. What speed can it gives to the mass?
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
Pls show the workings
2 answers