How would i do this problem.....i'm obviously doing something wrong.?

Question

A 100-kg block is pushed up a 30 degree incilne that is 10m long. If the coefficient of friction betweeen the block and the incline is 0.1, what is the constant force parallel to the incline that is required to move the block from rest at the bottom of the incline to the top in 3 seconds?

Okay; i am attempting to find Fn by using mgcos(theta)=(100)(9.81)cos(30)= 849.6N. Then i do f=(coeff. friction)(Fn)=(.1)(849.6)=84.95N.
Where am i going wrong here? I don't think i fully understand this problem at all. Please help me.

Answer 1

You need to find out what the acceleration of the block is along the plane that would make it take 3 seconds to travel 10m.

distance = a*t^2/2 = 10meters, plug in 3sec for time and we solve for a, which is 2.22meter/sec^2

ma = F - 0.1*mgCos(30)-mgSin(30)

100Kg*2.22m/sec^2 = F - 574.87
F = 796.87 newtons

please check for errors

Answer 2

Step 1. Solve for the acceleration using the formula,
s=ut +1/2at^2, where s is the distance travelled, u is the initial velocity (in this case u=0 because
the block starts from rest),
a is the acceleration in m/s/s, and t is the time in sec.

Now substitute the given values:

10=(0)3+1/2a(3)^2
10=0+4.5a
a=10/4.5
=2.22m/s/s

Step 2.

If the block were frictionless, then F=ma, but since friction acts on the block an added force has to be applied to overcome the force of friction which is given by the formula: f=uN, where f equals the force of friction in Newtons, u is the coefficient of friction, and N is the load acting perpendicular to the surface expressed in Newtons. Substituting the given values, we get:
F=100*2.22=222Newtons
f=0.1*(100*9.8*cos30)
=0.1(100*9.8*0.866)
=84.9Newtons
Add the two together,i.e. F+f=222+84.9=306.9Newtons

Answer 3

The calculation you have done tells you how much force is required just to overcome the friction (84.96 N).

Next you need to figure out how much force it will take to overcome the force of gravity. This is equal to the component of the gravitational force that is parallel to the incline.

Finally, you need to figure out how much force is required to accelerate the block from 0 m/s to 6.667 m/s in 3 seconds. Why 6.667 m/s? Because the average speed over the 10 m is 3.333 m/s (= 10 m / 3 sec). If it starts at 0 m/s and ends at 6.667 m/s, then it will average 3.333 m/s, as desired. You can use F = ma to solve for this third component of the required force; since you know m and a, you can simply multiply them to calculate F.

Answer 4

I think this is a power issue because you've been given a time. There is a constant resistance due to friction through the distance, and that's work done. Then there's the amount that the block has been raised, and that's energy supplied to the block. If you add that work and energy and divide it by the time involved, you should get the force you need. Feel free to b---h at me if I'm wrong because it's getting late.

Answer 5

Ff= 0.1(100)(9.80665)cos30
af = Ff/100 = 0.1*9.80665cos30 ("down" the incline)
s = s0 + v0t + (1/2)at^2, but a =(F-Ff)/m
s0 = 0, v0 = 0, so
s = (1/2)((F - mgsin30 - Ff)/m)t^2
F = (2ms/t^2) + Ff + mgsin30
F = (2ms/t^2) + μmgcos30 + mgsin30
F = m((2s/t^2) + g(μcos30 + sin30)
F = 100((2*10/3^2) + 9.80665(1+0.1√3)/2)
F = 100(20/9 + 9.80665(1+0.1√3)/2)
F = 797.4828 N
(84.928 N to overcome friction, 490.3325 N to overcome weight of the block acting "down" the plane, and 222.2222 N to privide acceleration)

Answer 6

Hey, I hope this helps. ur finding work is it?

W = Fd + μN
W = (100)(g)(10) + (0.1)(100)(g)(cos30)

u hv to add the normal working force and the extra force to overcome the friction. u didnt add them in ur attempt. but im not sure about the time issue because your question asked for force which has nothing to do with time. adding the time factor in would make it associated to finding power.