A 3.0 kg block is moved up a 37 degree incline under a constant horizontal force of 40.0 N. The coefficient of friction is 0.10 and the block is displaced 2.0 m up the incline. Calculate the change in the kinetic energy of the block.
Wb = m*g = 3kg * 9.8N/kg = 29.4 N. =
Weight of block.
Fp = 29.4*sin37 = 17.69 N. = Force
parallel to the incline.
Fn = 29.4*cos37 = 25.61 N. = Normal force = Force perpendicular to the
incline.
Fk = u*Fn = 0.1 * 25.61 = 2.56 N =
Fap = 40/cos37 = 50.09 N. Parallel to
incline.
a = (Fap-Fk)/m = (50.09-2.56)/3 = 15.84
m/s^2.
V^2 = Vo^2 + 2a*d = 0 + 2*15.84*2=63.4
V = 7.96 m/s.
KE=0.5m*V^2 = 0.5*3*63.4 = 95.1 Joules.
Correction:
a = (Fap-Fp-Fk)/m = (50.09-17.69-2.56)/3
= 9.95 m/s^2.
V^2 = Vo^2 + 2a*d = 0 + 2*9.95*2 = 39.8
V = 6.30 m/s.
KE = 0.5m*V^2 = 0.5*3*39.8 = 59.7 J.
To calculate the change in the kinetic energy of the block, we need to first determine the net work done on the block. The net work is equal to the change in kinetic energy.
Here's how we can find the net work done:
Step 1: Find the gravitational force acting on the block parallel to the incline.
- The gravitational force acting parallel to the incline is given by Fg_parallel = m * g * sin(θ), where
m = mass of the block = 3.0 kg (given),
g = acceleration due to gravity = 9.8 m/s^2, and
θ = angle of the incline = 37 degrees (given in the problem).
Plugging in the values, we get:
Fg_parallel = 3.0 kg * 9.8 m/s^2 * sin(37 degrees) ≈ 17.417 N
Step 2: Find the force of friction acting on the block.
- The force of friction is given by the equation F_friction = μ * N, where
μ = coefficient of friction = 0.10 (given in the problem), and
N = normal force.
The normal force can be calculated as N = m * g * cos(θ), where
m = mass of the block = 3.0 kg (given), and
g = acceleration due to gravity = 9.8 m/s^2.
Plugging in the values, we get:
N = 3.0 kg * 9.8 m/s^2 * cos(37 degrees) ≈ 23.868 N
Now, substituting the values in the equation for the force of friction:
F_friction = 0.10 * 23.868 N ≈ 2.387 N
Step 3: Find the force exerted by the person pushing the block up the incline.
- The force exerted by the person is given as 40.0 N (given in the problem).
Step 4: Calculate the net force acting on the block.
- The net force is given by the equation F_net = F_applied - F_friction - Fg_parallel, where
F_applied = force applied by the person = 40.0 N,
F_friction = force of friction = 2.387 N (from Step 2), and
Fg_parallel = gravitational force parallel to the incline = 17.417 N (from Step 1).
Substituting the values, we get:
F_net = 40.0 N - 2.387 N - 17.417 N ≈ 20.196 N
Step 5: Calculate the work done by the net force.
- The work done by the net force is given by the equation W_net = F_net * d * cos(θ), where
d = displacement of the block = 2.0 m (given in the problem),
F_net = net force acting on the block, and
θ = angle of the incline = 37 degrees (given in the problem).
Substituting the values, we get:
W_net = 20.196 N * 2.0 m * cos(37 degrees)
Using a scientific calculator or trigonometric table, calculate the cosine of 37 degrees and multiply it by 20.196 N * 2.0 m. Let's call this value W_net.
Step 6: Calculate the change in the kinetic energy of the block.
- The change in kinetic energy is equal to the net work done on the block.
Therefore, ΔKE (change in kinetic energy) = W_net.
Substituting the calculated value, you can determine the change in kinetic energy of the block.