Inuit Ed has entered his husky dog in the 100.0 m sled dog dash.

Question

Inuit Ed has entered his husky dog in the 100.0 m sled dog dash. The dog pulls the sled (of mass 113 kg) with its applied muscle force through thick snow which offers a resistance due to surface μ = 0. 231. Also, the dog must deal with the weight of Ed on the sled, a personal gravity force of 499 N {downward}.

At the start of the race, husky moves the sled, from rest , with an acceleration of 2.30 m/s2 {west} until halfway mark. The track is level so no motion exists in the vertical direction.

a) Find Ed’s mass

b) What normal force exists to keep the sled level?

c) Draw a FBD for the sled and Ed together during the first half of race

d) Find the applied force of husky muscle

e) How fast was husky moving at the 50.0 m mark?

f) How long did it take to reach the 50.0 m mark?

Since Ed wants husky to set a new race record and finish in less than 10.0 s, he decides to jump off the sled at the 50.0 m mark

g) which of Newton’s laws guarantees that this will give husky more acceleration? Describe the relation between mass and acceleration.

h) without Ed and with husky keeping his same applied force, calculate the new acceleration.

i) calculate exactly where husky will be, relative to start, when the clock hits 10.0s

Answer 1

Here is a sketch and a FBD:

The weight (Force of Gravity or F_G) = weight of the sled + weight of Ed = 113•9.81 + 499 = 1607.53 N

Because there is no motion in the y-direction (up and down), then the sum of the forces in the y-direction is zero. So F_N - F_G = 0 or F_N = F_G

The Force of Friction (F_F) is calculated using F_F = μ•F_N = 0.231•1607.53

The sum of the forces in the x-direction (∑F_x) equals the mass of the system (sled + Ed) times the acceleration:

∑F_x = ma

F_H - F_F = (m_sled + m_Ed)a (and we can say m_Ed = 499/9.81 kg)

That gives you enough to solve for the F_H.

To find out how hast they are going at the 50 meter mark, use the kinematic equation of motion:

v_f² = v_i² + 2ax where v_i = 0 (they start from rest), a (the acceleration) is given and x = 50. To find the time, you can use the kinematic equation v_f = v_i + a•t since you just solved for v_f.

Once he jumps off, the force of friction drops because F_G gets smaller meaning the acceleration gets larger (it will no longer be 2.30 m/s²) To find the acceleration use F_H - F_F = (m_sled)a with the new F_F but the old calculated F_H (this is Newton's second law).

For the last question, you'll need to divide the problem into two parts, the time it takes to go the first 50 m (which you calculated above) and then see what time is left (10 s minus the time for the first 50 meters) and then use the kinematic equations of motion to find "x" using the "new" acceleration and the time you just calculated.