A 2 kg object falls 3 meters how much potential energy did the object have beforw it fell
1 answer:
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Refer to the figure shown below.
g = 9.8 m/s², the acceleration due to gravity.
W = mg, the weight of the mug.
θ = 17°, the angle of the ramp.
Let μ = the coefficient of static friction.
The force acting down the ramp is
F = W sin θ = W sin(17°) = 0.2924W N
The normal reaction is
N = W cosθ = W cos(17°) = 0.9563W N
The resistive force due to friction is
R = μN = 0.9563μW N
For static equilibrium,
μN = F
0.9563μW =0.2924W
μ = 0.3058
The frictional force is F = μN = 0.2924W
The minimum value of μ required to prevent the mug from sliding satisfies
the condition
R > F
0.9563μW > 0.2924W
μ > 002924/.9563 = 0.306
Answer:
The frictional force is 0.2924mg, where m = the mass of the mug.
The minimum coefficient of static friction is 0.306
g = 9.8 m/s², the acceleration due to gravity.
W = mg, the weight of the mug.
θ = 17°, the angle of the ramp.
Let μ = the coefficient of static friction.
The force acting down the ramp is
F = W sin θ = W sin(17°) = 0.2924W N
The normal reaction is
N = W cosθ = W cos(17°) = 0.9563W N
The resistive force due to friction is
R = μN = 0.9563μW N
For static equilibrium,
μN = F
0.9563μW =0.2924W
μ = 0.3058
The frictional force is F = μN = 0.2924W
The minimum value of μ required to prevent the mug from sliding satisfies
the condition
R > F
0.9563μW > 0.2924W
μ > 002924/.9563 = 0.306
Answer:
The frictional force is 0.2924mg, where m = the mass of the mug.
The minimum coefficient of static friction is 0.306
Let the mass of 2500 kg car be and it's velocity be
and the mass of 1500 kg car be
and it's velocity be
.
After the bumping the mass be M and it's velocity be V.
By law of conservation of momentum we have
2500 * 5 + 1500 * 1=4000 * V
V = 14000/4000 = 7/2 = 3.5 m/s
So the velocity of the two-car train = 3.5 m/s
Answer:
Shawn's speed relative to Susan's speed = 10 mph
Resultant velocity = 82.32 mph
Explanation:
The given data :-
i) Susan driving in north and speed of Susan is ( v₁ ) = 53 mph.
ii) Shawn driving in east and speed of Shawn is ( v₂ ) = 63 mph.
iii) The speed of both Susan and Shawn is relative to earth.
iv) The angle between Susan in north and Shawn in east is 90°.
We have to find Shawn's speed relative to Susan's speed.
v₂₁ = v₂ - v₁ = 63 - 53 = 10 mph
Resultant velocity,
v = 82.32 mph