A 2.5-kg block is sliding along a rough horizontal surface and collides with a horizontal spring whose spring constant is 320 N/m. Unstretched, the spring is 20.0 cm long. The block causes the spring to compress to a length of 12.5 cm as the block temporarily comes to rest. The coefficient of kinetic friction between the block and the horizontal surface is 0.25. a) How much work is done by the spring as it brings the block to rest? b) How much work is done on the block by friction as the block is in contact with the spring? c) What was the speed of the block when it first came into contact with the spring?
mass, m = 2.5 kg
spring constant, k = 320 N/m
compression of spring, Δx = 0.075 m
Coefficient of kinetic friction, μk = 0.25
acceleration due to gravity, g = 9.8 m/s²
Let
initial velocity of block, u =u m/s
final velocity of block, v = 0 m/s
Work done by block, W
= loss of KE
= (1/2)mu²
Work done by friction, Wf
= μkmgΔx
Work done by spring, Ws
= (1/2)kΔx²
Equate total work:
W=Wf+Ws
(1/2)2.5u²=0.25*2.5*9.8*0.075+(1/2)*320*0.075²
Solve for u=3.157 m/s
To solve this problem, we can break it down into three parts:
a) Calculate the work done by the spring.
b) Calculate the work done on the block by friction.
c) Calculate the speed of the block when it first came into contact with the spring.
Let's begin with part (a):
a) The work done by the spring can be calculated using the formula for the potential energy stored in a spring:
U = (1/2)kx^2
Where U is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.
Given:
Spring constant, k = 320 N/m
Displacement, x = (Unstretched length - Compressed length) = (20 cm - 12.5 cm) = 7.5 cm = 0.075 m
Using the formula:
U = (1/2)kx^2
U = (1/2)(320 N/m)(0.075 m)^2
U = 0.9 J
Therefore, the work done by the spring as it brings the block to rest is 0.9 J.
Moving on to part (b):
b) The work done on the block by friction can be calculated using the formula:
W = f * d * cos(theta)
Where W is the work, f is the force of friction, d is the displacement, and theta is the angle between the force and displacement vectors.
The force of friction can be calculated using the formula:
f = u * N
Where u is the coefficient of kinetic friction, and N is the normal force.
Since the block is on a horizontal surface, the normal force is equal to the weight of the block (mg).
Given:
Coefficient of kinetic friction, u = 0.25
Mass of the block, m = 2.5 kg
Displacement, d = Compressed length of the spring = 12.5 cm = 0.125 m
Angle, theta = 180 degrees (force and displacement are opposite)
Calculating the force of friction:
f = u * N
f = 0.25 * mg
Calculating the work done by friction:
W = f * d * cos(theta)
W = (0.25 * mg) * 0.125 m * cos(180 degrees)
W = (-0.25 * mg * 0.125 m)
Therefore, the work done on the block by friction is (-0.25 * mg * 0.125 m) J.
Finally, let's move on to part (c):
c) The speed of the block when it first comes into contact with the spring can be calculated using the principle of conservation of mechanical energy:
KE_initial + PE_initial + W_external = KE_final + PE_final
Since the initial and final potential energies are both zero (as the block starts from rest and comes to rest), we can simplify the equation to:
KE_initial + W_external = KE_final
Given:
Work done by the spring, W_external = 0.9 J
Mass of the block, m = 2.5 kg
Simplifying the equation:
(1/2)mv_initial^2 + W_external = 0
(1/2)(2.5 kg)v_initial^2 + 0.9 J = 0
Solving for v_initial:
(1/2)(2.5 kg)v_initial^2 = -0.9 J
v_initial^2 = (-0.9 J) / (1.25 kg)
v_initial^2 = -0.72 m^2/s^2
Since the speed cannot be negative, we ignore the negative sign.
Taking the square root of both sides:
v_initial = sqrt(0.72 m^2/s^2)
Therefore, the speed of the block when it first came into contact with the spring is approximately 0.85 m/s.
To solve this problem, we need to use the principles of work and energy.
a) To find the amount of work done by the spring, we can use the formula for the work done by a spring:
Work = (1/2) * k * x^2
where k is the spring constant and x is the displacement of the spring from its equilibrium position.
In this case, the spring constant is given as 320 N/m, and the displacement is the difference in length between the unstretched and compressed positions of the spring, which is (20.0 cm - 12.5 cm) = 7.5 cm = 0.075 m.
Therefore, the work done by the spring is:
Work = (1/2) * (320 N/m) * (0.075 m)^2
= 0.9 J
b) To find the work done on the block by friction, we first need to find the force of friction. The force of friction can be calculated using the formula:
Friction = μ * N
where μ is the coefficient of kinetic friction and N is the normal force.
The normal force can be found using the equation:
N = m * g
where m is the mass of the block and g is the acceleration due to gravity (9.8 m/s^2).
In this case, the mass of the block is given as 2.5 kg. Therefore,
N = (2.5 kg) * (9.8 m/s^2)
= 24.5 N
The force of friction can now be calculated:
Friction = μ * N
= (0.25) * (24.5 N)
= 6.125 N
The work done by friction is given by the formula:
Work = Friction * d
where d is the displacement of the block on the rough surface. As the block comes to rest, the displacement of the block can be calculated as the difference in length from its initial position to the compressed position of the spring, which is (20.0 cm - 12.5 cm) = 7.5 cm = 0.075 m.
Therefore, the work done by friction is:
Work = (6.125 N) * (0.075 m)
= 0.459 J
c) To find the speed of the block when it first came into contact with the spring, we can use the principle of conservation of mechanical energy.
Initially, the block is moving, so it has kinetic energy given by:
Kinetic Energy = (1/2) * m * v^2
where m is the mass of the block and v is its initial velocity.
At the point of contact with the spring, the block temporarily comes to rest, so all of its initial kinetic energy is transferred to the spring as potential energy.
Therefore, we can set the initial kinetic energy equal to the potential energy stored in the spring:
(1/2) * m * v^2 = (1/2) * k * x^2
Substituting the given values, we get:
(1/2) * (2.5 kg) * v^2 = (1/2) * (320 N/m) * (0.075 m)^2
Simplifying the equation, we find:
v^2 = (320 N/m) * (0.075 m)^2 / (2.5 kg)
v^2 = 0.288 m^2/s^2
v ≈ 0.537 m/s
Therefore, the speed of the block when it first came into contact with the spring was approximately 0.537 m/s.