Question

Susan's 12.0 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30∘ above the floor. The tension is a constant 29.0 N and the coefficient of friction is 0.180. Use work and energy to find Paul's speed after being pulled 2.70m .Express your answer with the appropriate units.

Answer 1

Answer:1.71 m/s

Explanation:

Given

mass of Susan $m=12 kg$

Inclination $\theta =30^{\circ}$

Tension $T=29 N$

coefficient of Friction $\mu =0.18$

Resolving Forces Along x axis

$F_x=T\cos \theta -f_r$

where $f_r=friction\ Force$  

$F_y=mg-N-T\sin \theta$

since there is no movement in Y direction therefore

$N=mg-T\sin \theta$

and $f_r=\mu N$

Thus $F_x=T\cos \theta -\mu N$

$F_x=29\cos (30)-\0.18\times (12\times 9.8-29\sin (30))$                

$F_x=25.114-18.558$

$F_x=6.556 N$

Work done by applied Force is equal to change to kinetic Energy

$F_x\cdot x=\frac{1}{2}\cdot mv_f^2-\frac{1}{2}\cdot mv_i^2$

$6.556\times 2.7=\frac{1}{2}\cdot 12\times v_f^2$

$v_f^2=\frac{6.556\times 2.7\times 2}{12}$

$v_f^2=2.95$

$v_f=1.717 m/s$        

Answer 2

The speed depends on the displacement and time. The speed of Paul after being pulled is 1.71 m/s.

What is the velocity?

The velocity of an object is defined as the change in the position of the object with respect to time.

Given that the mass of Paul is 12 kg. The coefficient of friction is 0.180. The tension force is 29 N. The angle of inclination is 30 degrees.

The friction force is given as,

$F_r = \mu N$

Where $\mu$ is the coefficient of friction and N is the force along the y-axis.

The force along the x-axis is given as,

$F_x = Tcos\theta - F_r$

Where T is the tension force and $\theta$ is the angle of inclination.

The force along the y-axis is given as,

$F_y= mg-N-Tsin\theta$

Where m is the mass of Paul and g is the gravitational acceleration. There is no force applied in the vertical direction.

$0 = mg -N - Tsin\theta$

$N = mg - Tsin\theta$

Substituting the values, we get,

$N = 12\times 9.8 - 29 \times sin30^\circ$

$N = 103.1\;\rm N$

The friction force will be,

$F_r = 0.180 \times 103.1$

$F_r = 18.558\;\rm N$

The force along the x-axis is given as,

$F_x = 29 \times cos 30^\circ - 18.558$

$F_x = 6.556\;\rm N$

The work done by the force is equivalent to the change in the kinetic energy. This is given as below.

$F_x \times x = \dfrac {1}{2}mv^2$

Where x is the distance along the x-axis and v is the velocity of movement.

$6.556 \times 2.7 = \dfrac {1}{2}\times 12 \times v^2$

$v^2 = 2.95$

$v = 1.71 \;\rm m/s$

Hence we can conclude that the speed of Paul after being pulled is 1.71 m/s.

To know more about the velocity, follow the link given below.

https://brainly.com/question/862972.