Question by Zaveron J: DISTANCE?????????????????????
A block with mass m = 10.4 kg slides down an inclined plane of slope angle 26.0 ° with a constant velocity. It is then projected up the same plane with an initial speed 1.75 m/s. How far up the incline will the block move before coming to rest?

Best answer:

Answer by Prashant
Draw the free body diagram of the block while it slides down the incline : Since it slides with constance velocity, a = 0
mg acts downwards
Normal reaction, N acts perpendicular to the incline
Friction, f acts up the plane

So N = mg cos 26 …………….(1)
f = uN
and f = mg sin26
or uN = mg sin26………………(2)

From (1) and (2) ;
u = tan26 = 0.5

Now consider the upward motion of the block :
mg acts downwards
N acts perpendicular to the incline
f acts down the plane

Acceleration, a acts down the plane

N = mg cos26
f + mgsin26 = ma
or uN + mg sin26 = ma
or umg cos26 + mg sin26 = ma
or a = ug cos26 + g sin26 = 8.7 m/s^2

Now
Initial velocity, u = 1.75 m/s
a = – 8.7 m/s^2 (minus sign indicates that a and u are in the opposite direction)
Final velocity, v = 0
Use v^2 – u^2 = 2as
s = 17.6 cm

Hope this helps.

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by Kiran Raja Bahadur SRK

Question by Destiny L: distance???
A solid sphere of uniform density has a mass of 4.0×104 kg and a radius of 2.5 m. What is the gravitational force due to the sphere on a particle of mass 1.0 kg located at a distance of 3.00 m from the center of the sphere?

What is the gravitational force due to the sphere on a particle of mass 1.0 kg located at a distance of 1.25 m from the center of the sphere?

Best answer:

Answer by Nizar
F1= Gmm’/d^2 =6.67428 *[10^(- 11)]*4*(10 ^4)*1/9=2 .966*10^-7 N

F2= Gmm’/d’^2 =17.08 *10^-7 N

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