A skier weighing 86.2 kg starts from rest and slides down a 32.0-m frictionless slope that is inclined at an angle of 15.0° with
the horizontal. Ignore air resistance.
a. calculate the work done by gravity on the skier.
b.calculate the work done by the normal force on the skier
a. calculate the work done by gravity on the skier.
b.calculate the work done by the normal force on the skier
1 answer:
Answer:
A. = 26.11kJ
B = 0N
Explanation:
Mass = 86.2kg
Distance (s) = 32m
Angle of the slope = 15°
a. Work done due to gravity
W = force * distance * angle of displacement
W = F * S * cos θ
Force = mass * acceleration due to gravity
Force = 86.2 * 9.8
Force = 844.76N
Work = 844.76 * 32 * cos 15
Work = 26111.21J
Work = 26.11kJ
b. The workdone by the normal force is equal to zero since the normal force is perpendicular to the displacement.
Hence the normal force does not act on the body.
N = 0N
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Answer:
The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s
Explanation:
From Newton's second law, F = mg and also from coulomb's law F= Eq
Dividing both equations by mass;
F/m = Eq/m = mg/m, then
g = Eq/m --------equation 1
Again, in a projectile motion, the time of flight (T) is given as
T = (2usinθ/g) ---------equation 2
Substitute in the value of g into equation 2
Charge of proton = 1.6 X 10⁻¹⁹ C
Mass of proton = 1.67 X 10⁻²⁷ kg
E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°
Solving for T;
T = 7.83 X10⁻⁷ s