A sled of mass m is given a kick on a frozen pond. The kick gives it an initial

Question

A sled of mass m is given a kick on a frozen pond. The kick gives it an initial speed of 2.00 m/s. The coefficient of kinetic friction between the sled and the ice is 0.100: Using both work-energy theorem and force considerations, find the distance the sled moves before it stops. A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The pulling force is 100 N parallel to the incline, which makes an angle of 20.0 degree with the horizontal. The coefficient of kinetic friction is 0.400 and the crate is pulled 5.00 m along the incline. How much work is done by force of gravity? How much energy is lost because of friction? How much work is done by the applied 100-N force? What is the change in kinetic energy of the crate? What is the speed of the crate after it has been pulled 5.00 m? A tandem (two-person) bicycle team must overcome a force of 165 N to maintain a speed of 9.00 m/s. Find the power required per rider, assuming that each contributes equally. Express your answer in watts and in horsepower.

Explanation / Answer

3)

Let the distance moved be d

Now, the friction force on the the sled,

F = u*mg

where u = 0.1

So, acceleration, a = F/m = u*g

Using the equation of motion,

v^2 = u^2 +2as

where v = final speed = 0

u = 2 m/s

a = -u*g = -0.1*9.8 = -0.98 m/s2

s = d

So, 0 = 2^2 -2*0.98*d

So, d = 2.04 m <-------answer

Now using Work energy theorem,

Initial KE = 0.5*mu^2

u = 2 m/s

Now, the work done by friction, W = -umg*d

So, by workenergy theorem , KEi + W = 0

So, 0.5*m*2^2 - 0.1*m*9.8*d = 0

So, d = 2.04 m <---------answer

4)

a)

work done by gravity, Wg = -mg*h

here h = 5*sin(20 deg)

So,

Wg = -10*9.8*5*sin(20 deg) = -167.6 J

b)

Energy lost due to friction,

Ef = -u*mg*cos(20)*d

= -0.4*10*9.8*cos(20 deg)*5

= -184.2 J

c)

Work done by force,

Wf = 100*5 = 500 J

d)

Change in KE = Wf + Wg + Ef

= 500 - 167.6 - 184.2

= 148.2 J

e)

Now,

change in KE = net Work done

So,

0.5*10*(v^2 - 1.5^2) = 148.2

So, v = 5.65 m/s

5)

Power = F*v

So, P = 165*9 = 1485 W

So, P = 1485/746 = 1.99 hp

So, power required by each = P/2 = 1.99/2 = 1 hp

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