1. The cart is moved 2.0 cm beyond the equilibrium point( the spring is stretche
ID: 1462011 • Letter: 1
Question
1. The cart is moved 2.0 cm beyond the equilibrium point( the spring is stretched ).a) what is the elastic potential energy stored in the spring? Does the energy depend on the mass of the cart?
b) if the cart is released from rest at the 2.0 cm mark, what will its kinetic energy be when it first passes through the springs equilibrium position ?
c) if the cart is released from rest at the 2.0 cm mark, what will its velocity be when it first passes through the springs equilibrium position? 1. The cart is moved 2.0 cm beyond the equilibrium point( the spring is stretched ).
a) what is the elastic potential energy stored in the spring? Does the energy depend on the mass of the cart?
b) if the cart is released from rest at the 2.0 cm mark, what will its kinetic energy be when it first passes through the springs equilibrium position ?
c) if the cart is released from rest at the 2.0 cm mark, what will its velocity be when it first passes through the springs equilibrium position?
a) what is the elastic potential energy stored in the spring? Does the energy depend on the mass of the cart?
b) if the cart is released from rest at the 2.0 cm mark, what will its kinetic energy be when it first passes through the springs equilibrium position ?
c) if the cart is released from rest at the 2.0 cm mark, what will its velocity be when it first passes through the springs equilibrium position?
Explanation / Answer
a) Elastic Potential Energy, P.E. = 1/2*k*x2;
where 'k' is the spring constant; and 'x' is the elongation = 2.0 cm;
No, Energy doesnot depend on the mass of the spring.
b) K.E. = P.E.
K.E. = 1/2*k*x2 (J)
c) 1/2*mv2 = 1/2*k*x2
velocity, v = x*sqrt[k/m] (m/s)
where 'm' is the mass of the cart
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