A man with a mass of 100kg is running at a speed of 2.5 m/s. From that run, he jumps on a skateboard of mass of 3kg. Assuming no energy lost to friction. Determine the final velocity of the man on the board.
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A: Mass (m1)= 62.5kg. U1 = 3.8m/s Mass (m2) = 67.5kg. U2= 0m/s
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- A 65 kg person rides a 20 kg bicycle at a speed of 11 m/s. The cyclist runs into a parked car and comes to a fast stop. The bike collapses 0.95 m during the crash. a) Determine the kinetic energy of the cyclist/bike before the crash. b) Find the average force exerted on the cyclist/bike by the parked car.After coming down a slope, a 60-kg skier is coasting northward on a level, snowy surface at a constant 15 m/s. Her 5.0-kg cat, initially running southward at 3.8 m/s, leaps into her arms, and she catches it. (a) Determine the amount of kinetic energy converted to internal energy in the Earth reference frame. (b) What is the velocity, measured in the Earth reference frame, of an inertial reference frame in which the cat's kinetic energy does not change?..A 6.30-kg object is initially moving so that its x-component of velocity is 6.00 m/s and y-component of velocity is -1.60 m/s. (a) What is the kinetic energy of the object at this time? (b) Find the change in kinetic energy of the object it its velocity changes so that its new x-component is 8.50 m/s and its new y-component is 5.00 m/s.
- Tarzan (80 kg) drops from a height of 2.5 m to swing down and pick up Jane. If he picks up Jane (60 kg) with an inelastic collision, determine the following. The velocity just before picking up Jane. (7.1 m/s) The velocity just after picking up Jane. (4.04 m/s) The percentage of energy lost when picking up Jane. (43%) The height (h), after picking up Jane. (0.82 m)Two cars are moving. The first car has twice the mass of the second car, but only half as much kinetic energy. Then, both cars increase their speed by 8.00 m/s. They then have the same kinetic energy. The original speeds of the two cars are, respectively, 22.6 m/s, v2 = 11.3 m/s. (A) V1 = 5.66 m/s, v2 = 11.3 m/s. V1 = 14.1 m/s, v2 = 7.07 m/s. V1 = 11.3 m/s, v2 = 5.66 m/s.What is the speef of the bomb after the push? What about the speed of the astronaut after the push? How much energy did the astronaut expend pushing the bomb away?
- You push a box of mass 21.4 kg with your car up to an icy hill slope of irregular shape to a height 5.1 m. The box has a speed 13.3 m/s when it starts up the hill, the same time that you brake. It then rises up to the top (with no friction) to a flat area before elastically colliding with a smaller box of mass 14 kg. The boxes then fall off a sheer cliff individually to the ground (with no drag). (a) How far from the cliff does the heavier box hit the ground? (b) What is the direction of motion of the heavier box as it hits the ground? = ŷ + 2 (c) How far from the cliff does the lighter box hit the ground? (d) What is the direction of motion of the lighter box as it hits the ground? W = ŷ + 2A 0.015-kg bullet is fired into a stationary block of wood of mass of 5.2 kg. The relative speed of the bullet with respect to the wood is zero after collision (The bullet becomes embedded on the block of wood). If the speed of the bullet before collision is 462 m/s, what is the kinetic energy of the bullet plus-wood combination right after the collision? Assume no change in the potential energy of the system before and after the collision. Express your answer in two decimal places, in Joules.