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castortr0y [4]
3 years ago
12

A sled is moving at a constant speed down a surface inclined at 45.0 with the horizontal and travels 30 meters in 4 seconds calc

ulate the vertical velocity of the sled at the end of the 4 seconds

Physics
1 answer:
Alexxandr [17]3 years ago
3 0

Because the speed of the sled is constant throughout this ordeal, we know that the instantaneous and average velocities are equal, and the velocity of the sled is

v=\dfrac{-30\,\mathrm m}{4\,\mathrm s}=-7.5\,\dfrac{\mathrm m}{\mathrm s}

(negative because we're taking the downward direction to be negative)

The vertical component of this vector is given by

v_y=\|v\|\sin\theta

where \theta is the angle made by the velocity vector with the positive x-axis (taken to be the right direction). Here, \theta=225^\circ, so

v_y=\left(7.5\,\dfrac{\mathrm m}{\mathrm s}\right)\sin225^\circ\approx-5.3\,\dfrac{\mathrm m}{\mathrm s}

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v=3.73m/s,   a=3.35m/s^{2}

Explanation:

In order to solve this problem, we must first draw a diagram that will represent the situation (See attached picture).

So there are two ways in which we can solve this proble. One by using integrls and the other by using energy analysis. I'll do it by analyzing the energy of the system.

So first, we ned to know what the constant of the spring is, so we can find it by analyzing the energy on points A and C. So we get the following:

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We know that the kinetic energy in C will be zero because its speed is supposed to be zero. We also know the potential energy due to the gravity in C is also zero because we are at a height of 0m. So we can simplify the equation to get:

K_{A}+U_{Ag}=U_{Cs}+W_{Cf}

so now we can use the respective formulas for kinetic energy and potential energy, so we get:

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k=\frac{mV_{A}^{2}+mgh_{A}-fy_C}{y_{C}^{2}}

and now we can substitute the respective data:

k=\frac{(2000kg)(4m/s)^{2}+(2000kg)(9.8m/s^{2})(2m)-(17000N)(2m)}{(2m)^{2}}

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k= 9300N/m

Once we got the constant of the spring, we can now find the speed of the elevator at point B, which is one meter below the contact point, so we do the same analysis of energies, like this:

K_{A}+U_{Ag}=K_{B}+U_{Bs}+U_{Bg}+W_{Bf}

In this case ther are no zero values to eliminate so we work with all the types of energies involved in the equation, so we get:

\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}mV_{B}^{2}+\frac{1}{2}ky_{B}^{2}+mgh_{B}+fy_{B}

and we can solve for the Velocity in B, so we get:

V_{B}=\sqrt{\frac{mV_{A}^{2}-ky_{B}^{2}+2mgh_{B}-2fy_{B}}{m}}

we can now input all the data directly, so we get:

V_{B}=\sqrt{\frac{(2000kg)(4m/s)^{2}-(9300N/m)(1m)^{2}+2(2000kg)(9.8m/s^{2})(1m)-2(17000N)(1m)}{(2000kg)}}

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so we get:

F_{s}+f-W=ma

Supposing the acceleration is positive when it points upwards.

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a=\frac{F_{s}+f-W}{m}

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a=\frac{ky_{B}+f-W}{m}

and now we substitute the data we know:

a=\frac{(9300N/m)(1m)+(17000N)-(2000kg)(9.8m/s^{2})}{2000kg}

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3.35m/s^{2}

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