A block of mass m = 2.75 kg is attached to a spring of spring constant k = 550 N

Question

A block of mass m = 2.75 kg is attached to a spring of spring constant k = 550 N/m and is placed on a horizontal surface with one end of the spring fixed to a wall. The block is then pressed against the spring so that the spring compresses by an amount x = 0.15 m to the left from its equillibrium position, and the block is released from rest.

(a) Assuming that there is no friction, what would be the speed of the block as it passes the equillibrium position (x = 0) of the spring?

(b) What is the work done by the spring on the block in this period?

(c) If the coefficient of kinetic friction between the block and the horizontal surface is 0.20, obtain the speed of the block as it first passes the equillibrium position of the spring.

(d) In the case with friction, as in part (c), how far to the right of x = 0 will the block reach?

Explanation / Answer

a)

m = mass = 2.75 kg

k = 550 N/m

x = compression of spring = 0.15 m

let the speed be ''v''

using conservation of energy

kinetic energy = spring potential energy

(0.5) m v2 = (0.5) k x2

inserting the values

(0.5) (2.75) V2 = (0.5) (550) (0.15)2

V = 2.12 m/s

b)

work done by spring = kinetic energy = (0.5) m V2 = (0.5) (2.75) (2.12)2 = 6.18 J

c)

m = mass = 2.75 kg

k = 550 N/m

uk = coefficient of kinetic friction = 0.20

x = compression of spring = 0.15 m

friction force acting on block is given as :

Fk = uk mg

let the speed be ''v''

using conservation of energy

spring potential energy = kinetic energy + work done by friction

(0.5) k x2 = (0.5) m v2 + Fk x

(0.5) k x2 = (0.5) m v2 + uk mg x

inserting the values

(0.5) (550) (0.15)2 = (0.5) (2.75) v2 + (0.20) (2.75)(9.8)(0.15)

V = 1.98 m/s

d)

let the spring is stretched by ''d''

using conservation of energy

spring potential energy + work done by friction = kinetic energy

(0.5) k d2   + Fk d= (0.5) m v2

(0.5) k d2   + uk mg d = (0.5) m v2

inserting the values

(0.5)(550) d2  + (0.2) (2.75) (9.8) d = (0.5)(2.75) (1.98)2

d = 0.131 m

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