maths/physics question?

Question

Initially a small block of wood is at a point O on a rough plane inclined at15 degrees to the horizontal. The block is projected upwars with initial speed 4ms-1. The coefficient of riction between the block and the plane is 0.1. The block comes instantaneously to rest at A. Find the distance OA and the speed of the block as it passes through O on the way back down.

I have had some real trouble with this question. I know that you have to find the acceleration yet if the mass is not given how is this possible. Thanks a lot.

Answer 1

Start by putting everything you know about this problem down on paper.

Theta = 15 degrees, the incline of the plane. u = 4 m/sec the intial velocity upward along the plane. k = .1 the coefficient of friction. Now use some of this stuff to define some physics.

KE = 1/2 mu^2, the kinetic energy of the block when it gets goosed upward from point O. What do we know about energy...never gets created or destroyed. So what happens to KE...becomes PE, potential energy = mgh, and lost as friction energy Q = Fd = kNd; where h = d sin(theta) and d is that distance OA you're looking for. N = W cos(theta) = normal weight relative to the plane.

So KE = 1/2 mu^2 = mgh + kNd = mg d sin(theta) + kmgd cos(theta) = PE + Q and look what happens to m, the mass. Goes away in 1/2 u^2 = gd (sin(theta) + k cos(theta)). Now solve for d = (1/2) u^2/g[sin(theta) + k cos(theta)] and everything on the RHS is known; so you can do the math.

Now you can do pretty much the same thing in reverse to solve for v, the velocity through O as it comes back down. (By the way, it may NOT come back down as static friction is greater than sliding friction; so when the block stops at A, it could very well just stay there at A. You need to know what the static coefficient of friction is.)

So, coming back down pe = mgh = mg d sin(theta) = 1/2 mv^2 + Q = ke + friction energy, where pe is the potential energy at A after losing that friction energy Q = kNd = kd W cos(theta) going up, which is the same coming back down the plane. In other words, not all that pe is converted into ke; some of it goes up in smoke (heat), so to speak, as the block slides downward.

So we have pe = mg d sin(theta) = 1/2 mv^2 + kmg cos(theta) d; and, again, the m goes away. Yay! gd sin(thet) - kgd cos(theta) = 1/2 v^2 and v^2 = 2gd (sin(theta) - k cos(theta) so that v = sqrt(2gd (sin(theta) - k cos(theta)). Again, you have all the numbers (including the derived d) and you can do the math.

Note: When theta = 90 deg the v equation becomes v = sqrt(2gd), which is just the sqrt of the last term in v^2 = u^2 + 2gd, the well-known SUVAT equation often used in beginning physics. So if the plane is swung up vertically to 90 degrees incline, the block starting at u = 0 velocity, will give us v^2 = 2gd and v = sqrt(2gd) velocity as it drops d distance. Since N = 0 when theta = 90, there is no loss to friction heat.