Question

A box with mass 15.0 kg moves on a ramp that is inclined at an angle of 55.0∘ above the horizontal. The coefficient of kinetic friction between the box and the ramp surface is μk = 0.300.
a. Calculate the magnitude of the acceleration of the box if you push on the box with a constant force 190.0 N that is parallel to the ramp surface and directed down the ramp, moving the box down the ramp.
Express your answer with the appropriate units.
b. Calculate the magnitude of the acceleration of the box if you push on the box with a constant force 190.0 N that is parallel to the ramp surface and directed up the ramp, moving the box up the ramp.
Express your answer with the appropriate units.

Answer 1

Answer:

a) a = 19.0 m/s²

b) a = 2.9 m/s²

Explanation:

a) We draw the free body diagram of the box. There are 4 forces: the normal force N, the weight mg, the constant force F and the kinetic frictional force μ_kN. We can take the coordinate system which is rotated 55° from the horizontal, to ease the calculations. So, we write the equations of motion in each axis:

$x: F+mg\sin\theta-\mu_k N=ma\\y: N-mg\cos\theta=0 \implies N=mg\cos\theta$

Substituting the expression for N in the first equation, we have:

$F+mg\sin\theta-\mu_kmg\cos\theta=ma\\\\\implies a=\frac{F+mg\sin\theta-\mu_kmg\cos\theta}{m}$

If we plug in the given values, we have:

$a=\frac{190.0N+(15.0kg)(9.8m/s^{2})\sin55\°-0.300(15.0kg)(9.8m/s^{2})\cos55\° }{15.0kg} =19.0m/s^{2}$

Since we chose the right-downward direction as positive, the positive sign in this case means that the box is accelerated downwards above the ramp.

b) In this case, the constant force F and the kinetic frictional force μ_kN point to the opposite side. In other words, we can just only change the sign of this two forces in the equations of part (a) and obtain:

$a=\frac{-F+mg\sin\theta+\mu_kmg\cos\theta}{m}$

Plugging in the given values:

$a=\frac{-190.0N+(15.0kg)(9.8m/s^{2})\sin55\°+0.300(15.0kg)(9.8m/s^{2})\cos55\° }{15.0kg} =-2.9m/s^{2}$

Since we chose the right-downward direction as positive, the negative sign in this case means that the box is accelerated upwards above the ramp.

This means that the magnitude of the acceleration in this case is 2.9m/s².

Answer 2

A box with mass 15.0 kg moves on a ramp that is inclined at an angle of 55.0∘ above the horizontal has

• a. The magnitude of the acceleration when force directed down the ramp 19 meter per second squared.
• b.The magnitude of the acceleration when force directed up the ramp is -2.95 meter per second squared.

What is Newtons second law of motion?

Newtons second law of motion shows the relation between the force mass and acceleration of a body. It says, that the force applied on the body is equal to the product of mass of the body and the acceleration of it.

It can be given as,

$F=ma$

Here, (m) is the mass of the body and (a) is the acceleration.

It can be rewritten as,

$a=\dfrac{F}{m}$

Given information-

The mass of the box is 15 kg.

the angle of inclined surface is 55 degrees above the horizontal.

The coefficient of kinetic friction between the box and the ramp surface is 0.300.

The value of constant force is 190 N.

The forces acting on the box are-

• The normal force in this case is 190 N.

• The force parallel which try to accelerate the block is,

         $mg\sin \theta=15\times9.81\times\sin (55)\\mg\sin \theta=120.54 N$

• The frictional force-

         $F_f=0.3\times15\times9.81\times \cos (55)\\F_f=25.32$

• a. The magnitude of the acceleration when force directed down the ramp-

As the force directed down the ramp thus the friction force acts on the box is in negative direction. Thus.

$\sum F=190+120.41-25.32=285.09\rm N$

Thus the magnitude of the acceleration is,

$a=\dfrac{285.09}{15}\\a=19\rm m/s^2$

Thus, the magnitude of the acceleration when force directed down the ramp is 19 meter per second squared.

• b.The magnitude of the acceleration when force directed up the ramp-

As the force directed up the ramp thus the normal force acts on the box is in negative direction. Thus.

$\sum F=-190+120.41+25.32=-44.27\rm N$

Thus the magnitude of the acceleration is,

$a=\dfrac{-44.27}{15}\\a=-2.95\rm m/s^2$

Thus, the magnitude of the acceleration when force directed up the ramp is -2.95 meter per second squared.

• a. The magnitude of the acceleration when force directed down the ramp 19 meter per second squared.
• b.The magnitude of the acceleration when force directed up the ramp is -2.95 meter per second squared.

Learn more about the Newtons second law of motion here;

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