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Brums [2.3K]
3 years ago
7

How high has a 45 g golf ball been driven if it has a potential energy of 2.06 J?

Physics
2 answers:
Mice21 [21]3 years ago
7 0
PE=mgh
2.06 J = 2.06 kg·m²/s² = (0.045 kg)·(9.8 m/s²)·h
where h is the height the ball has reached
Solve for h:
h = 2.06/(0.045·9.8) = 4.67 to the nearest hundredth
h = 4.67 m

nydimaria [60]3 years ago
4 0
Here, U = mgh

We have: m = 45 g = 0.045 Kg
U = 2.06 J, g = 9.8 m/s², h = ?

Substitute the values in the formula:
2.06 = 0.045 × 9.8 × h
2.06 = 0.441 × h
h = 2.06 / 0.441
h = 4.67 m

In short, your final answer would be 4.67 meters

Hope this helps!

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A small block of mass m1 = 0.4 kg is placed on a long slab of mass m2 = 2.8 kg. Initially, the slab is stationary and the block
mel-nik [20]

Answer:

v₁ = 0.375 m / s ,   x = 0.335 m

Explanation:

Let's analyze this interesting exercise, the block moves and has a friction force with the tile, we assume that the speed of the block is constant, so the friction force opposes the block movement. For the only force that acts (action and reaction) this friction force exerted by the block that is in the direction of movement of the tile.

We can also see that the isolated system formed by the block and the tile will reach a stable speed where friction cannot give the system more energy, this speed can be found by treating the system with the conservation of linear momentum.

initial moment. Right at the start of the movement

       p₀ = m v₀ + 0

final moment. Just when it comes to equilibrium

      p_{f} = (m + M) v₁

how the forces are internal

       p₀ =p_{f}

       m v₀ = (m + M) v₁

       v₁ = m /m+M    v₀

let's calculate

       v₁ = 0.4 /(0.4 + 2.8)  3

       v₁ = 0.375 m / s

 

Let's apply Newton's second law to the Block, to find the friction force

Y axis

       N - W = 0

       N = W

       N = m g

where m is the mass of the block

the friction force has the formula

      fr = μ N

      fr = μ m g

We apply Newton's second law to slab    

X axis

       fr = M a

where M is the mass of the slab

       μ m g = M a

       a = μ g m / M

let's calculate

       a = 0.15  9.8  0.4 / 2.8

       a = 0.21 m / s²

With kinematics we can find the position

       v²= v₀²+2 a x

as the slab is initially at rest, its initial velocity is zero

       v² = 2 a x

       x = v2 / 2a

let's calculate

        x = 0.375²/2 0.21

        x = 0.335 m

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