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torisob [31]
3 years ago
5

3.3 kg block is on a perfectly smooth ramp that makes an angle of 52° with the horizontal. (a) What is the block's acceleration

(in m/s2) down the ramp? (Enter the magnitude.) m/s2 What is the force (in N) of the ramp on the block? (Enter the magnitude.) N (b) What force (in N) applied upward along and parallel to the ramp would allow the block to move with constant velocity?
Physics
1 answer:
Alexus [3.1K]3 years ago
3 0

Answer:

a)  a = 7.72 m / s²,  N = 19.9 N  and b)   F = 25.5 N

Explanation:

To solve this problem we will use Newton's second law, let's set a reference system with an axis parallel to the plane and gold perpendicular axis. Let's break down the weight (W)

    sin52 = Wx / W

    cos52 = Wy / W

    Wx = W sin52

    Wy = w cos 52

Let's write them equations

X axis

    Wx = ma

Y Axis

    N-Wy = 0

    N = Wy

a) Let's calculate the acceleration

    a = W sin52 / m = mg sin 52 / m

    a = g sin 52

    a = 9.8 sin52

    a = 7.72 m / s²

The force of the ramp is normal

    N = Wy = mg cos 52

    N = 3.3 9.8 cos 52

    N = 19.9 N

b) For the block to move at constant speed the sum of force on the axis must be zero,

    F - Wx = 0

    F = Wx

    F = mg sin52

   F = 3.3 9.8 sin 52

   F = 25.5 N

Parallel to the plane and going up

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(b) 0.100

For the block on the left, f_{k} =u_{k} n= 0.100(2.45N)=0.245N.

∑F_{x}=ma_{x}

–0.308N+0.245N=(0.250kg)a

a=−0.252m/s^{2} if the force of static friction is not too large.

For the block on the right, f_{k} =u_{k} n=0.490N. The maximum force of static friction would be larger, so no motion would begin, and the acceleration is zero

To learn more about kinetic energy, refer to:

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Static friction is what you are looking for.
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