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a 12 kg runaway grocery cart runs into a spring, attached to a wall, with spring constant 290 n/m and compresses it by 80 cm . part a what was the speed of the cart just before it hit the spring? neglect any friction.

Sagot :

The initial speed of the cart is  1.3m/s.

The velocity of an object is the rate of change of its position with respect to a given time interval.

Given

Mass m of the cart  = 12kg

Spring constant =290 N/m.

The compressed length of the spring  = 10 cm.

The initial kinetic energy of the cart is given below.

    K.E. = 1/2 m v2

    v =  initial velocity of the cart.

When the cart hits the spring, it is compressed by the cart. When the cart stops, due to the maximum compression, the kinetic energy of the cart will be zero. Hence its kinetic energy will be converted into elastic potential energy of the spring. This is given as,

U = 1/2 kx²

k = spring constant

x = compressed length of the spring.

This elastic potential energy of the spring will be equal to the initial kinetic energy of the cart.

U = 1/2 * 290*(0.80)² = 0.64.

U = 64J.

Elastic potential energy of the spring will be equal to the initial kinetic energy of the cart.

64 = (1/2) * 80 * v²

128 = 80 * v²

v² = 128/ 80

v² = 1.6

v = 1.26 ≅ 1.3

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