A 750-kg automobile is moving at 26.2 m/s at a height of 5.00 m above the bottom of a hill when it runs out of gasoline. The car coasts down

Question

A 750-kg automobile is moving at 26.2 m/s at a height of 5.00 m above the bottom of a hill when it runs out of gasoline. The car coasts down the hill and then continues coasting up the other side until it comes to rest. Ignoring frictional forces and air resistance, what is the value of h, the highest position the car reaches above the bottom of the hill?

in progress 0
Nho 4 years 2021-08-23T04:02:17+00:00 1 Answers 27 views 0

Answers ( )

  1. kimchi2
    0
    2021-08-23T04:03:39+00:00
    Reply

    To solve this problem it is necessary to apply to the concepts related to energy conservation. For this purpose we will consider potential energy and kinetic energy as the energies linked to the body. The final kinetic energy is null since everything is converted into potential energy, therefore

    Potential Energy can be defined as,

    PE = mgh

    Kinetic Energy can be defined as,

    K= \frac{1}{2} mv^2

    Now for Conservation of Energy,

    KE_i+PE_i = PE_f

    \frac{1}{2}mv_i^2+mgh_1 = mgh_2

    \frac{1}{2} (750kg) (26.2m/s)^2 + (750)(9.8)(5) = (750)(9.8)h_2

    h_2 = 40.0224m

    Therefore the highets position the car reaches above the bottom of the hill is 40.02m

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )