Which pair of graphs represent the same motion?

A child whose weight is 276 N slides down a 5.90 m playground slide that makes an angle of 34.0° with the horizontal. The coeffi

garik1379 [7]

Answer:

Explanation:

Frictional force acting on the child = μ mg cosθ

, μ is coefficient of kinetic friction , m is mass of child θ is inclination

work done by frictional force

μ mg cosθ x d , d is displacement on inclined plane

work done = .13 x 276 x cos34 x 5.9

= 175.5 J

This work will be converted into heat energy.

b ) Initial energy of child = mgh + 1/2 m v ² , h is height , v is initial velocity

= 276 x 5.9 sin34 + 1/2 x 276 / 9.8 x .518² [ mass m = 276 / g ]

= 910.59 + 3.77

= 914.36 J

loss of energy due to friction = 175.5

Net energy at the bottom

= 738.86 J

If v be the velocity at the bottom

1/2 m v² = 738 .86

.5 x (276 / 9.8) x v² = 738.86

v² = 52.47

v = 7.24 m /s .

4 0

3 years ago

Read 2 more answers

Winding a chain A 30-m-long chain hangs vertically from a cylinder attached to a winch. Assume there is no friction in the syste

Tcecarenko [31]

Answer:

Explanation:

Assuming that velocities remain negligible,

The work will be equal to the change in potential energy of the center of mass

mass is 30 m(5 kg/m) = 150 kg

W = 150(9.8)(30/2) = 22,050 J ≈ 22 kJ

5 0

2 years ago

Que es una variable independiente ?

nlexa [21]

Answer:

Variable independiente

Explanation:

una variable independiente. Es una variable que está sola y no es modificada por las medidas de otras variables

7 0

2 years ago

There is a bell at the top of a tower that is 95 m high. The mass of the bell is 21 kg. What is the potential energy of the bell

melomori [17]

Answer:

19,551 J!

Explanation:

The formula is PE = ham (h=height, a= acceleration or 9.8, m= mass)

PE = (95)(9.8)(21)

PE = 19,551 Joules

5 0

3 years ago

A 30.0 kg packing case is initially at rest on the floor of a 1500 kg pickup truck. The coefficient of static friction between t

alexgriva [62]

Answer:

(a). The magnitude of the frictional force acting on the case is 66 N northward.

(b). The magnitude of the friction force acting on the case is 58.8 N southward.

Explanation:

Given that,

Mass of packing case = 30.0 kg

Mass of truck = 1500 kg

Coefficient of static friction = 0.30

Coefficient of kinetic friction = 0.20

Acceleration = 2.20 m/s² northward

(A). We need to calculate the force acting on the case

Using formula of force

$F=ma$

Put the value into the formula

$F=30.0\times2.20$

$F=66\ N$

We need to calculate the maximum static frictional force

Using formula of frictional force

$f_{\mu}=\mu mg$

Put the value into the formula

$F_{\mu}=0.30\times30.0\times9.8$

$F_{\mu}=88.2$

The frictional force is greater then the force acting on the block

Then, The frictional force is 66 N northward.

(B). We need to calculate the force acting on the case

Using formula of force

$F = ma$

Put the value into the formula

$F=30.0\times3.40$

$F=102\ N$

The static friction force is less then the force action on the case

We need to calculate the maximum static frictional force

Using formula of frictional force

$f_{\mu}=\mu mg$

$f_{\mu}=0.20\times30.0\times9.8$

$f_{\mu}=58.8\ N$

So, The magnitude of the friction force acting on the case is 58.8 N southward.

Hence, This is the required solution.

7 0

3 years ago