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2 years ago

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A block of mass m slids up a ramp making an angle θ with the horizontal. The block has an initial KE of 1 2 m v2 0 as it starts

up the ramp and travels a distance L along the ramp before coming to momentary rest. v0 L How much work did kinetic friction do on the block between its starting point and the point it came to momentary rest?

1 answer:

Step2247 [10]2 years ago

5 0

Answer:

Explanation:

Given

mass of block is m

inclination of ramp is $\theta$

Initial kinetic energy $\frac{1}{2}mv_0^2$

Length of ramp is L

Block move a distance of L so it moves a vertical distance of $L\sin \theta$

Applying work Energy theorem i.e. change in kinetic energy of object is equal to work done by all the forces

Initial kinetic energy $K.E._i=\frac{1}{2}mv_0^2$

Final kinetic energy $K.E._f=0$

$K.E._i-K.E._f=Work\ done\ by\ kinetic\ friction+work\ done\ by\ gravity$

$\frac{1}{2}mv_0^2=W_f+W_g$

$W_f=\frac{1}{2}mv_0^2-mgL\sin \theta$

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Why does a balloon filled with the air from your lungs sink to the ground but a balloon filled with Helium floats ?

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Helium is lighter than air from the lungs.

Explanation:

A balloon filled with the air from lungs will sink to the ground but a balloon filled with Helium floats because helium is lighter than air.

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3 years ago

A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve

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Answer:

(a) v = 5.42m/s

(b) vo = 4.64m/s

(c) a = 2874.28m/s^2

(d) Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

$v=\sqrt{2gh}$ (1)

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h: height where the ball falls down = 1.50m

$v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}$

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

$h_{max}=\frac{v_o^2}{2g}$ (2)

vo: velocity of the ball where it starts its motion upward

You solve for vo and replace the values of the parameters:

$v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}$

The velocity of the ball is 4.64m/s

(c) The acceleration is given by:

$a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}$

$a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}$

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

$\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m$

THe compression of the ball when it strikes the floor is 5.11*10^-3m

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Physics question, answer completely with work

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