Ask question

Ask question

All categories

2 years ago

9

Magnus has reached the finals of a strength competition. In the first round, he has to pull a city bus as far as he can. One end

of a rope is attached to the bus and the other is tied around Magnus's waist. If a force gauge placed halfway down the rope reads out a constant 2750 2750 Newtons while Magnus pulls the bus a distance of 1.60 1.60 meters, how much work does the tension force do on Magnus? The rope is perfectly horizontal during the pull.

1 answer:

iragen [17]2 years ago

7 0

Answer:

The workdone is $W_d =-4400J$

Explanation:

The free body diagram is shown on the first uploaded image

From the question we are given that

The force is on the force gauge $F = 2750 N$

The distance that Magnus pulled the bus $d = 1.60m$

Generally the workdone by the tension force on Magnus is

$Workdone = Force * displacement \ in \ the \ direction \ of \ force$

$W_d = F * (-d)$

This negative sign show that is tension force is in the opposite direction to Magnus movement (i.e the movement of the bus )

Substituting value we have

$Workdone = - 2750 * 1.60$

$=-4400 J$

You might be interested in

Goal posts at the ends of football fields are padded as a safety measure for players who might run into them. How does thick pad

dimaraw [331]

When looking at it from a physics standpoint, the thick padding decreases the amount of force that a player experiences. Force equals mass times acceleration where acceleration is distance over time. The thick padding increases the amount of time the player makes contact with the post which decreases force.

8 0

2 years ago

Read 2 more answers

Compare these two collisions of a PE student with a wall.

Stolb23 [73]

1) The variable that is different in the two cases is $\Delta t$ , the duration of the collision

2) The change in momentum is the same in the two cases

3) The impulse is the same in the two cases

4) Case B will experience a greater force

Explanation:

1)

The variable that is different in the two cases is $\Delta t$ , the duration of the collision.

In fact, in the first case the wall is padded: this means that the collision will be "softer" and therefore will last longer, so the duration of the collision, $\Delta t$ , will be larger.

In the second case instead, the wall is unpadded: this means that the collision is "harder" and so it will last less time, therefore the duration of the collision $\Delta t$ will be smaller.

2)

The change in momentum in the two cases is the same.

In fact, the change in momentum is given by:

$\Delta p = m(v-u)$

where:

m is the mass of the student

u is the initial velocity

v is the final velocity

In both cases, we have:

m = 75 kg

u = 8 m/s

v = 0 (they both comes to rest)

Therefore, the change in momentum is

$\Delta p = (75)(0-8)=-600 kg m/s$

3)

The impulse in the two cases is the same.

In fact, impulse is defined as the product of force applied, F, and duration of the collision, $\Delta t$ :

$J=F \Delta t$

However, the force can be rewritten as product of mass (m) and acceleration (a), according to Newton's second law:

$F=ma$

So the impulse is

$J=ma\Delta t$

The acceleration can be rewritten as rate of change of velocity:

$a=\frac{\Delta v}{\Delta t}$

So the impulse becomes

$J=m\frac{\Delta v}{\Delta t}\Delta t = m\Delta v$

So, the impulse is equal to the change in momentum: and since in the two cases the change in momentum is the same, the impulse is the same as well.

4)

The force in the collision is related to the impulse by

$J=F\Delta t$

where

J is the impulse

F is the force

$\Delta t$ is the duration of the collision

The equation can be rewritten as

$F=\frac{J}{\Delta t}$

In the two situations described in the problem (A and B), we already said that the impulse is the same (because the change in momentum is the same). However, in case A (padded wall) the time $\Delta t$ is longer, while in case B (unpadded wall) the time $\Delta t$ is shorter: since the force F is inversely proportional to the duration of the collision, this means that in case B the student will experience a greatest force compared to case A.

Learn more about impulse:

brainly.com/question/9484203

#LearnwithBrainly

3 0

2 years ago

A boy of mass 80 kg slides down a vertical pole, and a frictional force of 480 N acts on him. What is his acceleration as he sli

olga_2 [115]

Answer:

His acceleration is $\overrightarrow{a}=4\frac{m}{s^{2}}$

Explanation:

Newton's second law states that acceleration of a body is cause by a net force, the relation between them is:

$\sum\overrightarrow{F}=m\overrightarrow{a}$

On the boy there're acting two forces, his weight (W) that points downward and the frictional force (f) that points upward (they boy moves downward and friction always is opposite to movement). So $\sum\overrightarrow{F}=\overrightarrow{W}+\overrightarrow{f}$ so (1) is:

$\overrightarrow{W}+\overrightarrow{f}=m\overrightarrow{a}$

Using the positive direction downward weight and gravitational acceleration(g) are positive and friction force is negative:

$W-f=m\overrightarrow{a}$ , solving for a:

$\overrightarrow{a}=\frac{W-f}{m}$ , weight is mg:

$\overrightarrow{a}=\frac{mg-f}{m}=\overrightarrow{f}=\frac{(80kg)(10\frac{m}{s^{2}})-(480)}{80}$

$\overrightarrow{a}=4\frac{m}{s^{2}}$

6 0

2 years ago

7. Imagine you are pushing a 15 kg cart full of 25 kg of bottled water up a 10o ramp. If the coefficient of friction is 0.02, wh

pentagon [3]

Answer:

The frictional force needed to overcome the cart is 4.83N

Explanation:

The frictional force can be obtained using the following formula:

$F= \mu R$

where $\mu$ is the coefficient of friction = 0.02

R = Normal reaction of the load = $mgcos\theta$ = $25 \times 9.81 \times cos 10$ = $241.52N$

Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.

$F=0.02 \times 241.52N$

F = 4.83 N

Hence, the frictional force needed to overcome the cart is 4.83N

4 0

2 years ago

An actor who memorizes his lines word-for-word has demonstrated _____ learning.

Nana76 [90]

<span>Mechanical association learning used by an actor to memorize his lines</span>

8 0

2 years ago

Read 2 more answers