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irakobra [83]
3 years ago
15

A mass of 6 kg with initial velocity 16 m/s travels through a wind tunnel that exerts a constant force 8 N for a distance 1.6 m.

It climbs a frictional incline of height 2.9 m inclined at an angle 16°, then moves along a second frictional surface of coefficient 0.1 before coming to rest.
The acceleration of gravity is 9.8 m/s^2. If the first frictional surface has a coefficient of 0.21 for a distance 1 m, how far does it slide along the second frictional region before coming to rest?
Physics
1 answer:
strojnjashka [21]3 years ago
6 0

Answer:

D=99.4665307m \approx 99.5m

Explanation:

From the question we are told that

Mass  m=6kg

Velocity of mass  V_m=16

Force of Tunnel  F_t=8N

Length of Tunnel L_t=1.6

Height of frictional incline H_i=2.9

Angle of inclination  \angle =16 \textdegree

Acceleration due to gravity  g=9.8m/s^2

First Frictional surface has a coefficient  \alpha_1 =0.21\ for\ d_c=1

Second Frictional surface has a coefficient \alpha _2=0.1

Generally the initial Kinetic energy is mathematically given by

K.E=\frac{1}{2}mv^2

K.E=\frac{1}{2}(6)(16)^2

K.E=768

Generally the work done by the Tunnel is mathematically given as

w_t=F_t*d_t

w_t=8*1.6

w_t=12.8J

Therefore

Total energy\ E_t=Initial\ kinetic energy\ K.E*Work done\ by\ tunnel\ W_t

E_t=K.E+E_t\\E_t=768J+12.8J

E_t=780.8J

Generally the energy lost while climbing is mathematically given as

E_c=mgh

E_c=(6)(9.8)(2.9)

E_c=170.52J

Generally the energy lost to friction is mathematically given as

E_f=\alpha *m*g*cos\textdegree*d_c

E_f=0.21*6*9.8*cos16*1

E_f=11.86965942 \approx 12J

Generally the energy left in the form of mass Em is mathematically given as

E_m=E_t+E_c+E_f

E_m=(768J)-(170.52)-(12)

E_m=585.48J

Since

E_m=\alpha_2*g*m*d

Therefore

It slide along the second frictional region

D=\frac{585.46}{0.1*9.81*6}

D=99.4665307m \approx 99.5m

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Answer:

(A) = 34,560 J

(B) Work done during the round trip is not zero, hence the resistive force is not a conservative force.

Explanation:

This question is incomplete, here is the complete question:

A girl rides her bike 5.4 km due east. While riding she experiences a resistive force from the air that has a magnitude of 3.1 N and points due west. She then turns around and rides due west, back to her starting point. The resistive force from the air on the return trip has a magnitude of 3.3 N and points due east.? (a) Find the work done by the resistive force during the round trip. (b) Based on your answer to part (a), is the resistive force a conservative force? Explain.

SOLUTION:

distance covered by the girl due east (De) = 5.4 km = 5400 m

magnitude of resistive force by the air (Fw) = 3.1 N due west

distance covered by the girl due west (Dw) = 5.4 km = 5400 m

magnitude of resistive force by the air (Fe) = 3.3 N due east

(A) to find the work done by the resistive force during the round trip we have to get the work done by the resistive force due west and add it to the work done by the resistive force due east

  • work done by the resistive force due west = (Fw.cosθ) x De

        where θ is the angle between the displacement and the force. The    

        displacement is due east while the force is due west, hence θ = 180°

       work done (W1) = (3.1 x cos 180) x 5400 = -16,740 j

  • work done by the resistive force due east = (Fe.cosθ) x Dw

        where θ is the angle between the displacement and the force. The    

        displacement is due west while the force is due east, hence θ = 180°

       work done (W2) = (3.3 x cos 180) x 5400 = -17,820 j

Hence work done during the round trip = W1 + W2 = (-16,740 ) + (-17,820)

= 34,560 J

(B) A conservative force is a force in which the work done by the force in moving an object around a closed path is zero. From part (A) above the work done during the round trip is not zero, hence the resistive force is not a conservative force.

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