Starting from the top and at rest a block slides down an inclined plane and has a velocity of v = 1.77 m/s ...

Question

at the bottom. The inclined plane makes a 28.8 deg. angle with the horizontal and is l=3.03 m long. The mass of the block is m=2.51 kg.

What is the coefficient of friction between the block and the plane?

Answer 1

To solve this problem, you need to use conservation of energy. At the beginning, the block is at rest at the top of this ramp. At the bottom of the ramp, the block still has some kinetic energy, but some of it has dissipated due to friction. Thus, the equation that governs this problem is:

Initial Potential Energy = Dissipated Energy due to Friction - Final Kinetic Energy

Since Potential Energy is defined only in the vertical direction, we must use the vertical components of the energy.

We can find the height of the ramp using trigonometry:

sin(theta) = height/incline -----------> height = incline*sin(theta)

height = 3.03*sin(28.8) = 1.46 m
Potential Energy = mgh = (2.51)(9.81)(1.46) = 35.95 J
Frictional Energy (vertical) = u*m*g*sin(theta)*h
Kinetic Energy (vertical) = 0.5*m*[v*sin(theta)]^2 = (0.5)(2.51)(1.77sin(28.8))^2

Kinetic Energy = 0.91 J

Plugging these values into the equation, you get:

35.95 = u*m*g*sin(theta)*h + 0.91

We have all the values, now solve for u

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Hope this helps

Answer 2

(a) Use the consistent acceleration formulation. The question tricks at that from the "consistent acceleration" area. in this question we can't use potential formulation's to help us because of the fact we at the instant are not given the peak of the prone airplane, yet in addition because of the fact the block is accelerating, and not vacationing at a relentless velocity. So we are given here concepts: u(preliminary velocity) = 0m/s (starts off from relax). v(very final velocity) = 0.60m/s (Take the action as being from the precise to the backside). s(displacement) = a million.6m (We take the diagonal to be the only measurement the block travels alongside). a = ? So we use the formulation v^2=u^2+2as (v^2-u^2)/2s = a nil.1125m/s/s. (b) For this question we ought to come to a sort forces. because of the fact the block is accelerating, there must be a internet tension directed down the airplane, we are asked to discover the coefficient of friction, so a frictional tension must be at artwork interior the alternative course. So resolving forces parallel to the airplane we get with trigonometry: F(internet) = mgcos(theta)-F(friction) F(friction)=-F(internet)+mgcos(theta) = -ma+mgcos(theta) =m(-a+gcos(theta) =2.a million(-0.1125+9.8cos(25)) = 18.42N. So F(friction) = uN = u(mgsin(theta)) (we come to a sort forces perpendicular to the airplane for N) u = F(friction)/(mgsin(theta) = 18.40 two/(2.a million*9.8sin(25)) = 2.12

Answer 3

The Potential Energy, PE, of the block at the top of the incline is converted into Work and Kinetic Energy at the bottom.

Given: mass of block, m=2.51kg
weight of block, W=mg=2.51*9.8
=24.6N
length of plane=3.03m
velocity at bottom,v=1.77m/s
angle of incline=28.8degrees

PE=Wh, where W is weight of block of 24.6N, and h the vertical displacement equal to 3.03sin28.8=3.03*0.482 =1.46m

PE=24.6*1.46=35.9joules

KE=1/2mv^2
=1/2*2.51*1.77^2
=3.93 Joules

Force of friction, f=uR, where u is the coefficient of friction and R the total force normal to the plane or Wcos28.8=
24.6*0.876=21.5N.

f=u*21.5

Component of W parallel to plane=Wsin28.8
=24.6*.482
=11.8N

Net force acting downwards, F=11.8-f
=11.8-u*21.5

Work done=F*l
=(11.8-21.5u)*3.03

PE=KE+Work

Substitute known values:

35.9=3.93+(11.8-21.5u)*3.03
35.9=3.93+35.8-65.1u
65.1u=3.83
u=3.83/65.1
=0.059

Answer 4

Friction, Fr=uN,
where u is the coeffcient we are trying to find and N is the normal reaction of the block on the ramp.

N=mg cos28.8
where m is the mass of block 2.51, g is 9.81.

the force that goes in the angle of 28.8 which serves to move the block i label as F, which is equal to mg sin28.8. so
F=mg sin 28.8

F-Fr=ma
where a is the acceleration in the angle 28.8 and F-Fr is the net force. using the kinematic equation v²=2as,
a=v²/2s
where v is 1.77, s is 3.03m
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to proceed with calculations,

1)Fr=F-ma
2)Fr=uN
3)uN=F-ma
4)u=(F-ma)N

u=0.490 to 3sf

Answer 5

U need to first calculate the acceleration, drawing free body diagram and resolving the forces horizontally(along the wedge) and vertically we get:
mgsin(x)-(mu)mgcos(x)=ma
or a=g(sin(x)-(mu)cos(x)
after calculating we can apply kinematics equation of motion
v^2=u^2+2as, where u=0 and s=l=3.03m and v=1.77m/s
from here (mu) can be easily calculated.