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elixir [45]
3 years ago
14

A man pushes his child in a grocery cart. The total mass of the cart and child is 30.0 kg. If the force of friction on the cart

is 15.0 N, how hard does the man have to push so that the cart accelerates at 1.50 m/s/s?
Physics
1 answer:
Ber [7]3 years ago
8 0
Newton's second law states that the resultant of the forces applied to an object is equal to the product between the object's mass and its acceleration:
\sum F = ma
where in our problem, m is the mass the (child+cart) and a is the acceleration of the system.

We are only concerned about what it happens on the horizontal axis, so there are two forces acting on the cart+child system: the force F of the man pushing it, and the frictional force F_f acting in the opposite direction. So Newton's second law can be rewritten as
F-F_a = ma
or
F=ma + F_f

since the frictional force is 15 N and we want to achieve an acceleration of a=1.50 m/s^2, we can substitute these values to find what is the force the man needs:
F=(30 kg)(1.5 m/s^2)+15 N=60 N
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3 years ago
A person pulls a crate of mass M = 63 kg a distance 40.0 m along a horizontal floor by a constant force FP = 130 N, which acts a
GrogVix [38]

Answer:

Check attachment for solution and diagrams

Explanation:

Given that,

Mass of crate m=63kg

Distance travelled d=40m

Horizontal force Fx=130N

Angle the force applied on cord makes with horizontal is θ=23°.

The weight of the crate is given by

W=mg

W=63×9.81

W=618.03N

Horizontal force Fx=130N

Resolving the applied Force F to the horizontal will give

Fx=FCos θ

F=Fx/Cos θ

F=130/Cos23

F=141.2N

a. Check attachment for model diagram

b. Check attachment for free body diagram

c. Check attachment for pictorial representation

d. Work done by gravitational force.

We, know that the body did not move upward, then the distance d=0

Work done is given as

W=F×d

So, d=0

W=F×0

W=0J

So, no work is done by gravity

e. Normal force?

Using newton law of motion

ΣFy = may

Since the body is not moving upward, then ay=0m/s²

N+141.2Sin23-618.03=0

N=618.03-141.2Sin23

N=562.86N

f. Work done by normal force.

The body is not moving upward, then the distance is zero

d=0

Work done by normal=normal force × distance

Wn=562.86×0

Wn=0J

No work is done by the normal force

g. Frictional force?

Since the coefficient of kinetic friction is zero, then the surface is frictionless

So, no frictional force is acting on the body

Fictional force is given as

Fr=μk•N

Given that, μk=0

Fr=0×562.86

Fr=0N

d. Work done by frictional force?

Since the frictional force Is zero, then, no work is done by friction

W(friction ) = frictional force × d

Here, the body moved a distance of 40m

W(fr)=0×40

W(fr)=0J

No work is done by friction

I. Work done by exerted force

The horizontal component of the exerted force is 130N and the body traveled a distance of 40m

Then, work done is given as

Workdone=force ×distance

Work done=130×40

W=5200J

W=5.2KJ

h. Net workdone?

Since no work is lost by friction, then, the net work done is equal to the work done by the exerted force.

Went, = work done by force exerted - work done by friction

Wnet=5200-0

Wnet, =5200

Wnet=5.2KJ

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3 years ago
Guys help me with this question.​
sleet_krkn [62]

Answer:

Explanation:

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1. The apparatus used are:

A is a source of light

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D is a microscope

2. The uses of the apparatus are:

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B - converges the rays of light from the source to the smoke cell.

C - is made of glass and used for encamping the smoke particles so as not to mix with air.

D - is used for the clear view or observation or study of the motion of the smoke particles in the cell.

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Answer:

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