A person on a sled coasts down a hill and then goes over a slight rise with speed 2.7 m/s. The top of the rise can be modeled as
a circle with a radius 4.1 m. The sled and occupant have a combined mass of 110 kg. If the coefficient of friction between the snow and the sled is 0.10, what friction force is exerted on the sled by the snow as the sled goes over the top of the rise?
It would be super helpful if you could explain how/why you came to your answer as well. Thanks!
It would be super helpful if you could explain how/why you came to your answer as well. Thanks!
1 answer:
Since we're dealing with radial acceleration around a circle, I used the radial acceleration equation a=v²/r. At the top of the hill, the force upward exerted by the hill is less than the weight of the sled. if v is large enough the term (g-v²/r) will become 0 and the sled will fly off the ground as it reaches the peak. Let me know if I can clarify any of my work.
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Answer:
28.1 m/s
Explanation:
= Initial velocity of the fish = 1.52 m/s
y = Height of the bird = 40 m
= Acceleration in y axis =
= Initial velocity in y axis = 0
The final velocity in x direction will remain the same as the initial velocity as there is no acceleration in the x direction
Resultant velocity is given by
The fish is moving at a velocity of 28.1 m/s when it hits the water.
Answer:
Speed of both blocks after collision is 2 m/s
Explanation:
It is given that,
Mass of both blocks, m₁ = m₂ = 1 kg
Velocity of first block, u₁ = 3 m/s
Velocity of other block, u₂ = 1 m/s
Since, both blocks stick after collision. So, it is a case of inelastic collision. The momentum remains conserved while the kinetic energy energy gets reduced after the collision. Let v is the common velocity of both blocks. Using the conservation of momentum as :
v = 2 m/s
Hence, their speed after collision is 2 m/s.