1) A block of mass M -0.250 kg is observed initially moving up a rough inclined

Question

1) A block of mass M -0.250 kg is observed initially moving up a rough inclined plane with a speed of vo -12 m/s. The plane is inclined an angle 25° to the horizontal as represented in the diagram below. The coefficient of kinetic friction is given by 0.400. Do the following: a) Draw an accurate free-body diagram with an appropriate coordinate system. Use Newton's second law to calculate the magnitude of the acceleration of the block with b) respect to the incline. block to come to For five extra points: calculate the time it would take for the credit instantaneous rest on the incline.

Explanation / Answer

initial velocity along the inclince=12 m/s

forces acting on the block are:

1. weight of the block=M*g=0.25*9.8=2.45 N, in vertically downward direction

2. normal force from the inclince=M*g*cos(theta)

=0.25*9.8*cos(25)=2.2205 N

it is perpendicular to the inclince.

3. friction force, downward on the inclince, opposing the motion.

magnitude=friction coefficient*normal force

=0.4*2.2205=0.8882 N

component of weight along the inclince=M*g*sin(theta)

=1.0354 N

so total force acting along the inclince in downward direction

=1.0354+0.8882=1.9236 N

acceleration opposing the initial velocity=force/mass

=1.9236/0.25=7.6944 m/s^2

part b:

when the block comes to instantaneous rest, its velocity will be 0.

using the formula:

final velocity=initial velocity+acceleration*time

==>0=12-7.6944*time

==>time=12/7.6944=1.5596 seconds

hence the time taken for the block to come to rest is 1.5596 seconds

Q2. part a;

average gravitational force=G*Mj*Me/r^2

=6.67*10^(-11)*1.9*10^27*4.8*10^22/(6.71*10^8)^2=1.3511*10^22 N

part b:

if average speed be v m/s then using the fact that

centripetal force is balancing gravitational pull,

Me*v^2/r=1.3511*10^22

==>v=sqrt(1.3511*10^22*6.71*10^8/(4.8*10^22))=1.3743*10^4 m/s

part c:

orbital period=total distance travelled in one complete revolution/speed

=2*pi*r/v

=2*pi*6.71*10^8/(1.3743*10^4)=3.0678*10^5 seconds

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