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13 hours ago

7

A 1000-kg car is moving along a straight road down a 30∘30∘ slope at a constant speed of 20.0m/s20.0m/s. What is the net force a

cting on the car?

1 answer:

Softa [913]12 hours ago

5 0

The overall force acting on the vehicle is zero

Explanation:

Let's evaluate the situation separately for the vertical direction and the horizontal direction along the slope.

Considering the direction perpendicular to the slope, two forces are in effect:

The weight component acting perpendicular to the slope, $mgcos \theta$ , directed into the slope
The normal force N, directed outward from the slope

Equilibrium exists here, indicating the net force in this direction is zero.

Now let’s examine the parallel direction to the slope. We have two forces present:

The weight component aligned with the slope, $mgsin \theta$ , directed down the slope
The frictional force $F_f$ , acting up the slope

The car moves at a constant speed in this direction, indicating that its acceleration is zero.

$a=0$

Thus, according to Newton's second law,

$F=ma$

implying the net force is zero:

$F=0$

Learn more about slopes and friction:

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Answer: $F=\frac{(M+m)g}{\mu _s}$

Explanation:

Provided:

The trolley, with mass M, is allowed to roll freely without friction.

The coefficient of friction between the trolley and mass m is $\mu _s$ .

A force F is applied to mass m.

The acceleration of the system is

$a=\frac{F}{M+m}$

The frictional force will counterbalance the weight of the block.

The frictional force is $=\mu _sN$

$N=ma$

$\mu _sN=mg$

$\mu _sma=mg$

$\mu _s=\frac{g}{a}$

$F=\frac{(M+m)g}{\mu _s}$

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If the coefficient of static friction between a table and a uniform massive rope is μs, what fraction of the rope can hang over

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Here's the procedure explained: Assume F represents the portion of the rope that is extending over the table. In this scenario, the frictional force that holds the rope on the table can be calculated using the formula: Ff = u*(1-f)*m*g. Additionally, it is important to determine the gravitational force that attempts to pull the rope off the table, Fg, calculated through: Fg = f*m*g. You then need to set these two equations equal to each other and resolve for f: f*m*g = u*(1-f)*m*g leads to f = u*(1-f) = u - uf. Simplifying gives f + uf = u, which results in f = u/(1+u) representing the fraction of the rope. This will lead you to the final answer.

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Softa [913]

Answer:

1)

$v_{oy}=11.29\ m/s$

2)

$y=7.39\ m$

Explanation:

Projectile Motion

When an object is projected near the surface of the Earth at an angle $\theta$ to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

The height of a projectile can be calculated using

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

where $y_o$ represents the initial height from ground level, $v_{oy}$ is the vertical component of the initial velocity, and t is the elapsed time.

The vertical speed component is expressed as

$v_y=v_{oy}-gt$

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of $v_o$ .

The peak height is attained when $v_y=0$ , which allows us to compute the time to reach that height.

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum height reached is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is equal to 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Continuing with the calculations for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Substituting known values yields

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

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