Ask question

Ask question

All categories

KonstantinChe [14]

2 years ago

8

A boy weighing 445 N swings on a 2-m long swing. If his horizontal speed at the lowest point is 3 m/s, what total force must the

ropes holding the swing be able to withstand?

1 answer:

levacccp [35]2 years ago

8 0

Explanation:

We need to calculate the centripetal force:

Fc = W + F

With Fc being the centripetal force, W the weight of the boy, F the centrifugal force (apparent).

We know that we can calculate the apparent centrifugal force thank to the formula:

F = (m·v²)/r = 204N

So we can write:

Fc = W + F = 445N + 204N = 649N

You might be interested in

A bird on a long migration flies 63 kilometers per hour for 2900

Fed [463]

Answer:

46

Explanation:

2900 divided by 63 (2900/63)

- 46.03174603

Rounded to the nearest whole number

is 46.

THE ANSWER IS 46

5 0

3 years ago

jasenka [17]

Answer:

answer on the google

Explanation:

6 0

3 years ago

Answer quick!!! (Edge 2021)

mihalych1998 [28]

Answer:

I believe the answer is B.

7 0

3 years ago

Read 2 more answers

1) Construct a graph showing the relationship between and independent variable and a dependent variable. 2) Demonstrate potentia

lana66690 [7]

Answer:

lolfsedddddddddddddd

Explanation:

8 0

3 years ago

Read 2 more answers

A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a forc

____ [38]

Answer:

Part a)

$a = 0.36 m/s^2$

Part b)

$a = -1.29 m/s^2$

Explanation:

Force applied by the student on the box is 80 N at an angle of 25 degree

so here two components of the force on the box is given as

$F_x = Fcos25$

$F_x = 80 cos25 = 72.5 N$

$F_y = Fsin25$

$F_y = 80 sin25 = 33.8 N$

now in vertical direction we can use force balance for the box to find the normal force on it

$F_n + F_y = mg$

$F_n = (25)(9.81) - 33.8$

$F_n = 211.45 N$

now kinetic friction on the box opposite to applied force due to rough floor is given as

$F_k = \mu F_n$

$F_k = (0.300)(211.45) = 63.44 N$

now the net force on the box in forward direction is given as

$F_{net} = F_x - F_k$

$F_{net} = 72.5 - 63.44 = 9.065 N$

now the acceleration of the box is given as

$a = \frac{F_{net}}{m}$

$a = \frac{9.065}{25} = 0.36 m/s^2$

Part b)

when box is pulled up along the inclined surface of angle 10 degree

now the two components of the force will be same along the inclined and perpendicular to inclined plane

$F_x = 72.5 N$

$F_y = 33.8 N$

now force balance perpendicular to inclined plane is given as

$F_n + F_y = mgcos\theta$

$F_n = (25)(9.81)cos10 - 33.8 = 207.7 N$

now the friction force opposite to the motion on the box is given as

$f_k = \mu F_n$

$f_k = (0.300)(207.7) = 62.3 N$

now the net pulling force along the inclined plane is given as

$F_{net} = F_x - F_k - mgsin10$

$F_{net} = 72.5 - 62.3 - (25)(9.81)sin10$

$F_{net} = -32.38 N$

now the box will decelerate and it is given as

$a = \frac{F_{net}}{m}$

$a = \frac{-32.38}{25} = -1.29 m/s^2$

7 0

3 years ago