A skier with a 65kg mass skies down a 30 degree incline hill. The coefficient of friction is 0.1.
1 answer:
Part a.
Refer to the diagram shown below.
m = 65 kg, the mass of the skier
θ = 30°, the angle of the incline
μ = 0.1, the coefficient of friction
W = mg , the weight of the skier
N = mg cosθ, the normal reaction
F = mg sinθ, the component of the weight acting down the hill
R = μN, the frictional force resisting motion down the plane
a = the acceleration down the plane.
The weight is
W = mg = (65 kg)*(9.8 m/s) = 637 N
Part b.
The normal force is
N = (637 N)*cos(30) = 551.658 N
The frictional force is
R = 0.1(551.658 N) = 55.1658 N
Part c.
Calculate the component of the weight acting down the hill.
F = mg sin(30) = (637 N)*sin(30) = 318.5 N
The equation of motion is
ma = F - R
(65 kg)*(a m/s²) = 318.5 - 55.1658 N
a = 263.3342/65 = 4.05 m/s²
Answers:
N = 551.7 N
a = 4.05 m/s²
Refer to the diagram shown below.
m = 65 kg, the mass of the skier
θ = 30°, the angle of the incline
μ = 0.1, the coefficient of friction
W = mg , the weight of the skier
N = mg cosθ, the normal reaction
F = mg sinθ, the component of the weight acting down the hill
R = μN, the frictional force resisting motion down the plane
a = the acceleration down the plane.
The weight is
W = mg = (65 kg)*(9.8 m/s) = 637 N
Part b.
The normal force is
N = (637 N)*cos(30) = 551.658 N
The frictional force is
R = 0.1(551.658 N) = 55.1658 N
Part c.
Calculate the component of the weight acting down the hill.
F = mg sin(30) = (637 N)*sin(30) = 318.5 N
The equation of motion is
ma = F - R
(65 kg)*(a m/s²) = 318.5 - 55.1658 N
a = 263.3342/65 = 4.05 m/s²
Answers:
N = 551.7 N
a = 4.05 m/s²
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The tension in the first and second rope are; 147 Newton and 98 Newton respectively.
Given the data in the question
- Mass of first block;
- Mass of second block,
- Tension on first rope;
- Tension on second rope;
To find the Tension in each of the ropes, we make use of the equation from Newton's Second Laws of Motion:
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For the Second Rope
Since only the block of mass 10kg is hang from the second, the tension in the second rope will be;
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Answer:
<h3>The answer is 2.51 s</h3>Explanation:
The time taken can be found by using the formula
d is the distance
v is the velocity
From the question we have
We have the final answer as
<h3>2.51 s</h3>Hope this helps you