The diagram below represents a bicyclist at the top of a hill, with four points labeled W, X, Y, and Z. Assume that

Physics

1 answer:

KiRa [710]4 years ago

8 0

Answer:

Explanation:

Potential energy is energy that is built up and saved. When you are moving and letting out that energy, that is kinetic energy. You would be using the least kinetic energy when you aren't moving, which is at the top of the hill because you move faster the more you go down.

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square root A 1400 kg car is coasting on a horizontal road with a speed of 18 m/s . After passing over an unpaved, sandy stretch

Lina20 [59]

Answer:

The net force on the car is 2560 N.

Explanation:

According to work energy theorem, the amount of work done is equal to the change of kinetic energy by an object. If ' $W$ ' be the work done on an object to change its kinetic energy from an initial value ' $K_{i}$ ' to the final value ' $K_{f}$ ', then mathematically,

$W = K_{f} - K_{i} = \dfrac{1}{2}~m~(v_{f}^{2} - v_{i}^{2})........................................(I)$

where ' $m$ ' is the mass of the object and ' $v_{i}$ ' and ' $v_{f}$ ' be the initial and final velocity of the object respectively. If ' $F_{net}$ ' be the net force applied on the car, as per given problem, and ' $s$ ' is the displacement occurs then we can write,

$W = F_{net}~.~s.......................................................(II)$

Given, $m = 1400~Kg,~v_{i} = 18~m~s^{-1}~v_{f} = 14~m~s^{-1}~and~s = 35~m$ .

Equating equations (I) and (II),

$&& - F_{net} \times 35~m = \dfrac{1}{2} \times 1400~Kg~\times(14^{2} - 18^{2})~m^{2}~s^{-2}\\&or,& F_{net} = \dfrac{\dfrac{1}{2} \times 1400~Kg~\times(14^{2} - 18^{2})}{35}~N\\&or,& F_{net} = 2560~N$