Physics, 11.10.2019 16:30 williamvh2001

A2.0 kg wood block is launched up a wooden ramp that is inclinedat a 35* angle. the block’s initial speed is 10m/s. (giventhat μk = 0.20)

a. what vertical height does the block reach above its startingpoint?
b. what speed does it have when it slides back down to its startingpoint?

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Answer from: crhockey08crhockey08

a) 4.0 m

We can solve this part by writing the equations of motion along the two directions: perpendicular to the slope and parallel to the slope.

Perpendicular to the slope:

$N-mgcos \theta =0$ (1)

where N is the normal reaction, m = 2.0 kg is the mass of the block, g = 9.8 m/s^2 is the acceleration of gravity, $\theta=35^{\circ}$ is the angle.

Parallel to the slope:

$-\mu_k N -mgsin \theta = ma$ (2)

where $\mu_k=0.20$ is the coefficient of friction, and a the acceleration, and where we have chosen up the slope as positive direction, so both forces are negative.

From (1) we get

$N=mg cos \theta$

And substituting into (2), we can find the acceleration:

$-\mu_k mg cos \theta -mgsin \theta = ma\\a=-\mu_k g cos\theta - g sin \theta = -(0.20)(9.8)(cos 35^{\circ})-(9.8)(sin 35^{\circ})=-7.2 m/s^2$

where the negative sign means the direction is down the slope.

Now we can find the distance travelled along the slope by using the SUVAT equation

v^2-u^2=2ad

where

v = 0 is the velocity when the block comes to rest

u = 10 m/s is the initial velocity

d is the distance travelled along the slope

Solving for d,

$d=\frac{v^2-u^2}{2a}=\frac{0-(10)^2}{2(-7.2)}=6.9 m$

And so, the vertical heigth gained by the block is

$h=d sin \theta = (6.9)(sin 35^{\circ})=4.0 m$

b) 7.4 m/s

The equation of motion along the direction parallel to the slope in this case is

$-\mu_k N +mgsin \theta = ma$

where this time we have taken down the slope as positive direction, so the component of the weight is positive while the frictional force is negative since the block slides downward (while friction acts upward). Solving for a, we find the new acceleration:

$a=-\mu_k g cos \theta + g sin \theta = -(0.20)(9.8)(cos 35^{\circ})+(9.8)(sin 35^{\circ})=4.0 m/s^2$

Now we can use again the SUVAT equation

v^2-u^2=2ad

where

v is the final velocity

u = 0 is the initial velocity

d = 6.9 m is the distance travelled along the slope

a = 4.0 m/s^2 is the acceleration

Solving for v,

$v=\sqrt{u^2+2ad}=\sqrt{0+2(4.0)(6.9)}=7.4 m/s$

Answer from: Quest

Ok what do you need with

Answer from: Quest

The figure is missing but we can answer anyway.the correct answer is "releasing the string". in fact, when the person is holding the bow and the arrow, the kinetic energy of the arrow is still zero (because the arrow is not moving). the moment the arrow is released, it acquires a velocity v, so its kinetic energy will become different from zero: $k= \frac{1}{2}mv^2$ where m is the mass of the arrow and v its velocity. all the other options, instead, do not increase the kinetic energy of the system: - pulling farther back on the string : this will just increase the elastic potential energy of the system- holding the arrow higher: this will increase the gravitational potential energy of the system (which is given by u=mgh, where h is the height above the ground)- decreasing the mass of the arrow: this will decrease the gravitational potential energy of the system (which is given by u=mgh, where m is the mass)

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