Question

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?

Answer 1

`D=99.4665307m \approx 99.5m`

Step-by-step 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=(1)/(2)mv^2`

`K.E=(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=(585.46)/(0.1*9.81*6)`

`D=99.4665307m \approx 99.5m`