A block of mass m = 4.88 kg is attached to a spring which is resting on a horizo

Question

A block of mass m = 4.88 kg is attached to a spring which is resting on a horizontal frictionless table. The block is pushed into the spring, compressing it by 5.00 m, and is then released from rest. The spring begins to push the block back toward the equilibrium position at x O m. The graph shows the component of the force (in N) exerted by the spring on the block versus the position of the block (in m relative to equilibrium. Use the graph to answer the following questions. How much work is done by the spring in pushing the block from its initial position at x _-5.00 m to x = 2.20 m? F(N) Number 4 3 x(m) What is the speed of the block when it reaches x 2.20 m? -5 3 -2 4 Number m/s What is the maximum speed of the block? Number m/s

Explanation / Answer

Given

mass m = 4.88 kg ,

From the given graph  

the work done is the area under the curve  

from the graph the x = -5 m to x = 0 m

the elastic potential energy U = 0.5*k*x^2 = area of the triangle

area of the triangle is A = 0.5*b*h = 0.5*5*6 J = 15 J

the work done = 0.5*k*x^2

15 = 0.5*k*5^2

k = 1.2 N/m

the maximum velocity is 0.5*k*x^2 = 0.5*m*v^2

v^2 = (k/m)(x^2)

v = sqrt((k/m)(x^2))

v = sqrt((1.2/4.88)(5^2)) m/s

v = 2.48 m/s

the work done by the spring from x = -5.0 m to x = 2.2 m

is W = 15+0.5*2.2*2.5 = 17.75 J

the speed of the block whtn it reaches x = 2.2 m is  

by conservation of energy  

0.5*m*v^2_ mean = 0.5*m*v^2 at (2.2m) + 0.5*k*x^2 at(2.2m)

0.5*4.88*2.48^2 = 0.5*4.88*v^2 + 0.5*1.2*2.2^2

v(2.2) = 2.23 m/s

the maximum speed of the block will be at mean position v = 2.48 m/s

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