Question

Coyote (m_c = 14.0 kg) is about to try out his new rocket sled invention (m_t = 45.0 kg), which can produce a thrust of 1600 N. 

 Part A: What is the normal force (Fs) exerted upwards by the road? 

 Part B: If the effective frictional coefficient between the rolling wheels and the road is μ = 0.050 what is the value of the frictional force (f)

 Part C: What will be the value of the acceleration (a) while Coyote and the apparatus are in motion with the rocket firing? 

Answer 1

Probi mc me = 14 ng Mr = as ng Mr Thret Thrust = 1600N. Q. Normal force by Road & Total weight Newton's 3rd Low] (Mr + mc) g FN- (14+45) x 9.81 FN=578.99 N b. friction coefficient t = 0.050 Then. friction force f = l. x Normal force. faux FW = MXFN = lux gordita fa 0.050x 518.99 f. = 28.94 N

ji Thrust a 16oon. (forward force) friction » 28.94 N (backward force) Mytme Thrust 1600N. P-2 f= 28.94N S. Fret ( 1600 – 28.94) N. by Newton's and low. fuet ma a Fret meotal 1600-28.94 1571.06 a = 14 + 45 (Met my a = 26.628 m/s2

Answer 2

The total normal force by the ground is equal to the weight of the body ( rocket + cyotee) by newton's third law. If friction coefficient is given we can easily find the friction force by multiplying it to Normal force.

Acceleration can be found by newton's 2nd law. Please find the attached solution.

Probi mc me = 14 ng Mr = as ng Mr Thret Thrust = 1600N. Q. Normal force by Road & Total weight Newton's 3rd Low] (Mr + mc) g FN- (14+45) x 9.81 FN=578.99 N b. friction coefficient t = 0.050 Then. friction force f = l. x Normal force. faux FW = MXFN = lux gordita fa 0.050x 518.99 f. = 28.94 N

ji Thrust a 16oon. (forward force) friction » 28.94 N (backward force) Mytme Thrust 1600N. P-2 f= 28.94N S. Fret ( 1600 – 28.94) N. by Newton's and low. fuet ma a Fret meotal 1600-28.94 1571.06 a = 14 + 45 (Met my a = 26.628 m/s2