All categories

10 months ago

I need help with a physics worksheet about Newtons Laws.

Physics

1 answer:

Viktor [21]10 months ago

8 0

The diagram of the object on the inclined plane is shown below

The mass of 7kg is exerted on the plane surface. It is acted upon by gravity. The force exerted on the surface is calculated by applying the formula,

F = mass x acceleration due to gravity

Assumme acceleration due to gravity is 9.8m/s^2, then

Force = 7 x 9.8 = 68.6N

The surface exerts an opposite force of the same magnitude with the force of the object but in opposite direction. Since it is inclined at an angle of 36.9 degrees,

Normal force = mgCosθ = 68.6Cos36.9 = 54.86N

Recall, frictional force = normal reaction x coefficient of friction

Given that coefficient of kinetic friction = 0.35,

Frictional force = 54.86 x 0.35 = 19.201N

This is the force that must be overcome to keep the object moving.

The force pulling the object upwards along the inclined plane is

mgSinθ = 7 x 9.8 x Sin36.9 = 41.19N

Since the velocity is constant, it means that there is no acceleration. The net force is zero. The force required to pull the mass and make it move at constant velocity must be equal to the sum of the exerted force and the the frictional force(Since it involves coeffricient of kinetic friction, it would cause the object to keep moving)

Thus,

Required force = 19.2 + 41.19 = 60.4N

Option E is correct

You might be interested in

What is kinetic energy?

Lemur [1.5K]

Answer:

A, the energy an object has due to its motion.

Explanation:

Kinetic energy is the energy created by motion.

3 0

2 years ago

Read 2 more answers

What is the main determining factor in defining boundaries between layers of earth's atmosphere?

konstantin123 [22]

The main determining factor in defining boundaries between layers of earth's atmosphere would be temperature changes in these layers. Temperature is one essential property that varies in the atmosphere. Based from this variation, the atmosphere is divided into four major layers and further to three smaller layers - troposphere, tropopause, the stratosphere, stratopause, the mesosphere, mesopause, and the thermosphere.The troposphere is the layer that is nearest to the surface of the Earth. It is the part where humans, plants and animals survive. Also, it is the warmest layer of the atmosphere. And as we go higher the atmosphere, the temperature would drop making it much cooler.

5 0

3 years ago

Read 2 more answers

A diverging lens (f1 = -12.0cm) is located 50.0 cm to the left of a converging lens (f2 = 34.0 cm). A 2.0 cm-tall object stands

pishuonlain [190]

Answer:

The final image relative to the converging lens is 34 cm.

Explanation:

Given that,

Focal length of diverging lens = -12.0 cm

Focal length of converging lens = 34.0 cm

Height of object = 2.0 cm

Distance of object = 12 cm

Because object at focal point

We need to calculate the image distance of diverging lens

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{-12}-\dfrac{1}{-12}$

$v=\infty$

The rays are parallel to the principle axis after passing from the diverging lens.

We need to calculate the image distance of converging lens

Now, object distance is ∞

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{34}-\dfrac{1}{\infty}$

$v=34$

The image distance is 34 cm right to the converging lens.

Hence, The final image relative to the converging lens is 34 cm.

5 0

3 years ago

A car of mass m accelerates from speed v_1 to speed v_2 while going up a slope that makes an angle theta with the horizontal. Th

Karo-lina-s [1.5K]

Answer:

Work done by external force is given as

$Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

Explanation:

As per work energy Theorem we can say that work done by all force on the car is equal to change in kinetic energy of the car

so we will have

$Work_{external} + Work_{gravity} + Work_{friction} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

now we have

$W_{gravity} = -mg(Lsin\theta)$

$W_{friction} = -\mu mgcos(\theta) L$

so from above equation

$Work_{external} - mgLsin\theta - \mu mgLcos(\theta) = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

so from above equation work done by external force is given as

$Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

8 0

3 years ago

An object of mass m is lowered at constant velocity at the end of a string of negligible mass. As it is lowered a vertical dista

bezimeni [28]

Answer:

Mechanical would have been conserved if only the force of gravity (the weight of the object does work on the system). The tension force does work also on the system but negative work instead. The net force acting of the system is zero since the upward tension in the string suspending the object is equal to the weight of the object but acting in the opposite direction. As a result they cancel out. In the equation above the effect of the tension force on the object has been neglected or not taken into consideration. For the mechanical energy E to be conserved, the work done by this tension force must be included into the equation. Otherwise it would seem as though energy has been generated in some manner that is equal in magnitude to the work done by the tension force.

The conserved form of the equation is given by

E = K + Ug + Wother.

In this case Wother = work done by the tension force.

In that form the total mechanical energy is conserved.

5 0

3 years ago