Suppose two carts, one twice as massive as the other, fly apart when the

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

7 0

3 years ago

A student says that a speed of 50 m/s is faster than a speed of 140 km/h because the number is bigger. What would you say to the

Yuri [45]

if you convert these into miles per hour 50 m/s would be higher, since

50 m/s = 111.85 mph and

140 km/h = 86.99 mph

3 0

3 years ago

4

Zigmanuir [339]

(a) The object moves with uniform velocity from A to B.

(b) The object moves with constant velocity from B to C.

(c) The object moves with increasing velocity from C to D.

<h3>Velocity of the object from point A to B</h3>

V(A to B) = (6 - 0)/(4 - 0) = 1.5 m/s

<h3>Velocity of the object from point B to C</h3>

V(B to C) = (6 - 6)/(11 - 4) = 0 m/s

<h3>Velocity of the object from point C to D</h3>

V(C to D) = (7 - 6)/(12 - 11) = 1 m/s

final velocity = 1 + 1.5 m/s = 2.5 m/s

Thus, we can conclude the following;

The object moves with uniform velocity from A to B.

The object moves with constant velocity from B to C.

The object moves with increasing velocity from C to D.

Learn more about velocity here: brainly.com/question/6504879

#SPJ1

4 0

2 years ago

A student wearing a frictionless roller skates on a horizontal is pushed by a friend with a constant force of 55N. How far must

umka21 [38]

Answer:

6.58m

Explanation:

The kinetic energy = Workdone on the roller

Workdone = Force * distance

Given

KE = Workdone = 362J

Force = 55N

Required

Distance

Substitute into the formula;

Workdone = Force * distance

362 = 55d

d = 362/55

d = 6.58m

Hence the student must push at a distance of 6.58m

3 0

3 years ago

A force acts on a 9.90 kg mobile object that moves from an initial position of to a final position of in 5.40 s. Find (a) the wo

horrorfan [7]

Given that,

Mass of object = 9.90 kg

Time =5.40 s

Suppose the force is (2.00i + 9.00j + 5.30k) N, initial position is (2.70i - 2.90j + 5.50k) m and final position is (-4.10i + 3.30j + 5.40k) m.

We need to calculate the displacement

Using formula of displacement

$s=r_{2}-r_{1}$

Where, $r_{1}$ = initial position

$r_{2}$ = final position

Put the value into the formula

$s= (-4.10i + 3.30j + 5.40k)-(2.70i - 2.90j + 5.50k)$

$s= -6.80i+6.20j-0.1k$

(a). We need to calculate the work done on the object

Using formula of work done

$W=F\cdot s$

Put the value into the formula

$W=(2.00i + 9.00j + 5.30k)\cdot (-6.80i+6.20j-0.1k)$

$W=-13.6+55.8-0.53$

$W=41.67\ J$

(b). We need to calculate the average power due to the force during that interval

Using formula of power

$P=\dfrac{W}{t}$

Where, P = power

W = work

t = time

Put the value into the formula

$P=\dfrac{41.67}{5.40}$

$P=7.71\ Watt$

(c). We need to calculate the angle between vectors

Using formula of angle

$\theta=\cos^{-1}(\dfrac{r_{1}r_{2}}{|r_{1}||r_{2}|})$

Put the value into the formula

$\theta=\cos^{-1}\dfrac{(-4.10i + 3.30j + 5.40k)\cdot(2.70i - 2.90j + 5.50k)}{7.54\times6.778})$

$\theta=79.7^{\circ}$

Hence, (a). The work done on the object by the force in the 5.40 s interval is 41.67 J.

(b). The average power due to the force during that interval is 7.71 Watt.

(c). The angle between vectors is 79.7°

7 0

3 years ago