Collage or College? Please clarify the School Subject:
Sra
"A 2.5-kg block slides down a 25 degree inclined plane with constant acceleration. The block starts from rest at the top. At the bottom, its velocity reaches 0.65 m/s. The length of the incline is 1.6m.
a) What is the acceleration of the block?
b) What is the coefficient of friction between the plane and the block?
c) Does the result of either (a) or (b) depend on the mass of the block?
4 answers
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First, quantify the given data in terms of symbols:
Length of inclined plane, S = 1.6m
inclination, θ =25°
mass of block, m = 2.5 kg
Initial velocity, u = 0 m/s
Final velocity, v = 0.65 m/s
acceleration = a m/s² (along incline)
coefficient of kinetic friction = μ
Look up your class notes or textbook and be familiar with the formulas required in kinematics of uniform accelerations, and inclined planes.
The approach could be:
1. calculate the net acceleration using the formula
2aS = v²-u²
2. Draw a free body diagram of the block, indication the direction and magnitude of the following forces:
weight due to gravity = mg (downwards)
Normal component of weight = mg*cos(θ) normal to inclined plane
downward component of weight F2= mg*sin(θ) (parallel to inclined plane, downwards)
frictional resistance, Fk= μN (in direction opposite to F2)
Therefore F=F2-Fk is the net force causing the calculated acceleration.
By equating F=ma, μ can be solved for.
Length of inclined plane, S = 1.6m
inclination, θ =25°
mass of block, m = 2.5 kg
Initial velocity, u = 0 m/s
Final velocity, v = 0.65 m/s
acceleration = a m/s² (along incline)
coefficient of kinetic friction = μ
Look up your class notes or textbook and be familiar with the formulas required in kinematics of uniform accelerations, and inclined planes.
The approach could be:
1. calculate the net acceleration using the formula
2aS = v²-u²
2. Draw a free body diagram of the block, indication the direction and magnitude of the following forces:
weight due to gravity = mg (downwards)
Normal component of weight = mg*cos(θ) normal to inclined plane
downward component of weight F2= mg*sin(θ) (parallel to inclined plane, downwards)
frictional resistance, Fk= μN (in direction opposite to F2)
Therefore F=F2-Fk is the net force causing the calculated acceleration.
By equating F=ma, μ can be solved for.
M = 2.5kg. = Mass of block.
A = 25o.
Vo = 0 = Initial velocity.
V = 0.65 m/s. = Final velocity.
d = 1.6m = Length of ramp.
M*g = 2.5*9.8 = 24.5 = Wt. of block.
Fp = 24.5*sin25 = 10.57 N. = Force parallel to incline.
Fn = 24.5*cos25 = 22.20 N. = Normal force.
Fk = u*Fn = 22.2u. = Force of kinetic friction.
a. V^2 = Vo^2 + 2a*d.
0.65^2 = 0 + 2a*1.6,
a = 0.132 m/s^2.
b. Fp-Fk = M*a.
10.57-22.2u = 2.5*0.132,
u = ?
c. Yes. Part a and b depend on mass of the block.
A = 25o.
Vo = 0 = Initial velocity.
V = 0.65 m/s. = Final velocity.
d = 1.6m = Length of ramp.
M*g = 2.5*9.8 = 24.5 = Wt. of block.
Fp = 24.5*sin25 = 10.57 N. = Force parallel to incline.
Fn = 24.5*cos25 = 22.20 N. = Normal force.
Fk = u*Fn = 22.2u. = Force of kinetic friction.
a. V^2 = Vo^2 + 2a*d.
0.65^2 = 0 + 2a*1.6,
a = 0.132 m/s^2.
b. Fp-Fk = M*a.
10.57-22.2u = 2.5*0.132,
u = ?
c. Yes. Part a and b depend on mass of the block.
1 answer