On a playground slide, a child has potential energy that decreases by 1000 J while her kinetic energy increases by only 900 J. How much energy is missing and where did it go?
A 18 kg child slides down a 4.0 m-high playground slide. She starts from rest, and her speed at the bottom is 2.8 m/s. What is the change in the thermal energy of the slide and the seat of her pants?
A 17.0 kg child travels down a playground slide 3.50 m high and reaches the bottom with a speed of 2.50 m/s. How much heat was generated due to friction as a result of this process? my teacher hadn't taught anything about heat..but he did give us this question, thus, i got no idea with this question. please help. thanks a lot. The heat generated equals the amount of (kinetic + potential) energy lost. That is because of the law of conservation of energy. In this case, M g H = 17x9.8x3.5 = 583.1 J of potential energy is lost while the kinetic energy increases by (1/2)x(17)x(2.5)^2 = 53.1 J. The rest of the potential energy loss, 530 J, is converted to heat
21.7 kg child descends a slide 3.5 m high and reaches the bottom with a speed of 2.2 m/s How many joules of thermal energy are created due to friction as a kid slides down a slide?
A child slides down a slide with a 24 incline, and at the bottom her speed is precisely half what it would have been if the slide had been frictionless.Calculate the coefficient of kinetic friction between the slide and the child.
At a playground, an 18 kg child plays on a slide that drops through a height of 2 m. The child starts at the top of the slide. Is the momentum conserved as she slidesdown the slide?
How large is the normal force of the slide on the child?
A 22 kg girl slides down a playground slide that is 3.6 m high. When she reaches the bottom of the slide, her speed is 1.7 m/s. The acceleration of gravity is 9.81 m/s2. How much energy was dissipated by fric-tion? Answer in units of J
A 34.0 kg childslides down a long slide in a playground. She starts from rest at a height h1 of 17.00 m. When she is partway down the slide, at a height h2 of 5.00 m, she ismoving at a speed of 9.40 m/s. Calculate the mechanical energy lost due to friction (as heat, etc.).
A 40 kg child is standing on the edge of a merry-go-round in a playground. Before they were deemed too dangerous, these were quite common. They were just huge rotating platforms you could sit on while someone spun you around in uniform circular motion until you screamed in terror (or possibly got sick). At any rate, let us assume the merry- go-round is a rotating disk of radius R = 1.5m and inertia M = 100kg and the child is a point particle located at the edge of the disk. The merry-go-round is initially at rest. The child’s brother tosses him a basketball from across the playground. Assume the basketball has a mass of 0.62kg and is thrown with a velocity of 20m/s perpendicular to the radius of the merry-go-round at the point where the child catches it. (a)[7 pt(s) ]Assume that there is no friction in the axle of the merry-go-round, so it can rotate freely. Assume that there is a lot of friction between the child and the merry-go-round, so that he does not slide as he catches the ball. What is the final angular velocity of the merry-go-round after the child catches the ball? (b)[4 pt(s) ]How much work is done on the merry-go-round? What does this work? (c)[3 pt(s) ]Draw the work-energy diagram to express the change in energy of the system consisting of only the merry-go-round. (d)[4 pt(s) ]Does the merry-go-round do any work on the ball? On the child? How much? (e)[3 pt(s) ]What is the centripetal force acting on the ball after the child catches it?
A child slides down a slide with a 21 incline, and atthe bottom her speed is preciselyhalf what it would have been if the slide had beenfrictionless.Calculate the coefficient of kinetic friction between theslide and the child.
A 32.0 kg child descends a slide 4.00 m high. She reaches the bottom with a speed of 2.40 m/s, was the mechanical energy conserved? Explain your reasoning and identify the energy transformations involved.