# A 3.50kg box is moving to the right with speed 8.00m/s on a horizontal, frictionless surface. At t = 0 a horizontal force is applied to the box. The force is directed to the left and has magnitude F(t)=( 6.00N/s2 )t2 Part A What distance does the box move from its position at t=0 before its speed is reduced to zero? Part B If the force continues to be applied, what is the velocity of the box at 3.50s ? Express your answer with the appropriate units.

###### Question:

Part A

What distance does the box move from its position at t=0 before its speed is reduced to zero?

Part B

If the force continues to be applied, what is the velocity of the box at 3.50s ?

Express your answer with the appropriate units.

## Answers

(a) The distance traveled by the box before stopping is 3.44 m.(b) The velocity of the box at 3.5 s is 40.55 m/s.The given parameters;mass of the box, m = 3.5 kgspeed of the box, v = 8 m/sforce applied to the box, F = (6 N/s²)t²The force applied on the box is given by Newton's second law of motion;[tex]F = ma\\\\F = \frac{mv}{t} \\\\\frac{dF}{dt} = mv\\\\F(t) = 6t^2\\\\\frac{dF}{dt} = 12 t\\\\12t = mv\\\\t = \frac{mv}{12} \\\\t = \frac{3.5 \times 8}{12} \\\\t = 2.33 \ s[/tex]The force applied to the box is calculated as;F = 6t²F = 6(2.33)²F = 32.57 NThe acceleration of the box is calculated as follows;[tex]a = \frac{F}{m} \\\\a = \frac{32.57}{3.5} \\\\a = 9.3 \ m/s^2[/tex]The distance traveled by the box before stopping is calculated as;[tex]v^2 = u^2 + 2as\\\\0 = u^2 + 2as\\\\-2as = u^2\\\\- s = \frac{u^2}{2a} \\\\-s = \frac{(8)^2}{2(9.3)} \\\\-s = 3.44 \ m\\\\|s| = 3.44 \ m[/tex]The velocity of the box at 3.5 s is calculated as;[tex]v_f = v_0+ at\\\\v_f = 8 + (9.3 \times 3.5)\\\\v_f = 40.55 \ m/s[/tex]Learn more here:https://brainly.com/question/18801986

Answer:F(t) = (-6.00 N/s^2)t^2 a(t) = (-3.00 N/s^2)t^2 Because F = ma, the acceleration function is the force function divided by mass (3.50 kg). Because the force is acting to the left, a negative has been introduced. Take the integral of the acceleration function with the power rule for integrals. Initial velocity is 8.00 m/s ∫a(t) dt=v(t)+v1v(t)=(-1m/s^4)*t^3+9 m/sSetting velocity equal to zero and solving for t. v(t)=0t^3=9s^3t=∛9s =2.08 sThe integral of velocity is position. The object begins at the origin so initial position is 0 ∫v(t) dt= x(t)x(t)=(-0.25m/s^4)*t^4+(9m/s)*tPlugging the t from step 3 into the x(t) function from step 4. This is the answer to part a. x(2.08)=14 mplug 3.50 s into the velocity function from step 2. Speed is the absolute value of velocity. This is the answer to part b. v(3.5)=(1 m/s^4)(3.5 s)^3+9 m/s = -18 m/sspeed(3.5 s)=║v(3.5)║=18 m/s