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Inga [223]
4 years ago
12

A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 joules (with the poten

tial energy equal to zero at ground level) and is moving upward with a kinetic energy of 50 joules. Air friction is negligible. The maximum height reached by the ball is most nearly
Physics
1 answer:
satela [25.4K]4 years ago
6 0

Answer: 20m

Explanation:

We will solve this question by applying the law of conservation of energy which states that the sum of potential energy and kinetic energy is always the same.

The PE is 0 at surface and maximum at top while the KE is maximum at surface and 0 at top.

From the question,

PE = mgh = 50 J -(1)

mg* 10 = 50

mg = 50/10

mg = 5

The total energy at that point = PE + KE = 50 + 50 = 100 J

Therefore, at topmost point, the PE will be 100 J

mgH = 100J , H is the needed height

Using the value of mg obtained above, we have

H= 100/5

H = 20 m

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7 0
3 years ago
A 98.0 N grocery cart is pushed 12.0 m along an aisle by a shopper who exerts a constant horizontal force of 40.0 N. If all fric
Digiron [165]

Answer:

9.8 m/s

Explanation:

The work done by the force pushing the cart is equal to the kinetic energy gained by the cart:

W=K_f -K_i

where

W is the work done

K_f is the final kinetic energy of the cart

K_i is the initial kinetic energy of the cart, which is zero because the cart starts from rest, so we can write:

W=K_f

But the work is equal to the product between the pushing force F and the displacement, so

W=Fd=(40.0 N)(12.0 m)=480 J

So, the final kinetic energy of the cart is 480 J. The formula for the kinetic energy is

K_f=\frac{1}{2}mv^2 (1)

where m is the mass of the cart and v its final speed.

We can find the mass because we know the weight of the cart, 98.0 N:

m=\frac{F_g}{g}=\frac{98.0 N}{9.8 m/s^2}=10 kg

Therefore, we can now re-arrange eq.(1) to find the final speed of the cart:

v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(480 J)}{10 kg}}=9.8 m/s



7 0
3 years ago
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