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Nutka1998 [239]
3 years ago
13

A grocery cart with a mass of 16 kg is pushed at constant speed along an aisle by a force fp = 12 n which acts at an angle of 17

° below the horizontal. find the work done by each of the forces on the cart if the aisle is 13 m long. work done by the applied force
Physics
2 answers:
olga_2 [115]3 years ago
6 0

Solution for this problem is we will be using the following formula:

Work done by applied force) = Force (applied) x s x cos θ

Where:

Force applied = 12 N

s = 13 m long

Angle = 17

So plugging in the values:


=>W (a) = 12 x 13 x cos 17

= 12 x 12.43196182

= 149.1835 J 

Eddi Din [679]3 years ago
4 0

Work done by the applied force : <u>Wp = 149,184 J </u>

<h3>Further explanation</h3>

Work is the transfer of energy caused by the force acting on a moving object

Work is the product of force with the displacement of objects.

Can be formulated

<h3>W = F x S </h3>

W = Work, J, Nm

F = Force, N

S = distance, m

Force and displacement of objects are vector . Multiplication of both is in accordance with the dot product concept of two vectors. Whereas W is a scalar quantity. Then if there is an angle formed by the force F and the displacement S is θ, then it can be formulated

W = F cos θ.S

There are 4 forces on the cart , so there are 4 work that occurs

  • work done by the applied force = Wp

= Fp.cos θ. S

= 12. cos 17. 13

Wp = 149,184 J

 

  • work done by the frictional force = Wf

the magnitude is the same as the force applied by the applied force (Fp) but the opposite direction = -149,184 J

  •  work done by the normal force = Wn

the direction is upward with an angle of θ = 90°, with F = mg

Wn = Fn.cos 90. S

Wn = mg cos 90. 13

Wn = 0

work done by the gravitational force = Wg

Wg = Fg. cos 90. S.

Wg = mg cos 90. 13

Wg = 0

<h3>Learn more </h3>

Isaac Newton's investigations of gravity

brainly.com/question/1747622

Gravitational force

brainly.com/question/7955425

increase the gravitational force between two objects

brainly.com/question/2306824

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The acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?
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We have that for the Question "the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?"

  • it can be said that the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line

From the question we are told

the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?

Generally the equation for the Force  is mathematically given as

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now

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Therefore

the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line

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A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a rope that goes over an ideal pulley. If the masses are
irga5000 [103]

Answer:

The time taken will be 0.553 seconds.

Explanation:

We should start off by finding the force exerted by the rope on the 3kg weight in this case.

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Weight of 3kg mass = 3 * 9.81 = 29.43 N

The force acting upward on the 3kg mass will equal the weight of the 5kg mass. Thus the resultant force acting on the 3kg mass is:

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We can now find the acceleration:

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We now use the following equation of motion to get the time taken to travel 1 meter:

s=u*t+\frac{1}{2} (a*t^2)

1=0*t+\frac{1}{2} (6.54*t^2)

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