A box of books weighing 229 N is shoved across the floor by a force of 420 N exerted downward at an angle of 35° below the horizontal.If µk between the box and the floor is 0.75, how long does it take to move the box 7 m, starting from rest? (If the box will not move, enter 0.)

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TAke the downward component of the force, add that to weight, to find the normal force. Then you can calculate friction force.

Compute the horizontal component of the pushing force.

horizontal force-friction=mass*acceleration

find acceleration.

distance=1/2 a t^2 solve for time.

To determine the time it takes to move the box, we need to consider the forces acting on the box and apply Newton's second law of motion.

1. Resolve the 420 N force into components:
- The vertical component is 420 N * sin(35°) = 240.4 N (downward).
- The horizontal component is 420 N * cos(35°) = 343.1 N (rightward).

2. Calculate the force of friction (Ffr) between the box and the floor using the formula:
Ffr = μk * FN
where:
- μk is the coefficient of kinetic friction (given as 0.75).
- FN is the normal force, which is equal to the weight of the box (229 N) acting perpendicular to the floor. Therefore, FN = 229 N.

Substituting these values, we get:
Ffr = 0.75 * 229 N = 171.75 N.

3. Determine the net horizontal force (Fnet) acting on the box:
Fnet = 343.1 N - Ffr
Fnet = 343.1 N - 171.75 N = 171.35 N (rightward).

Note: Since the box is moving horizontally, only the horizontal forces are considered.

4. Apply Newton's second law of motion to find the acceleration (a):
Fnet = ma
171.35 N = (mass of the box) * a

Rearranging the equation, we have:
a = 171.35 N / (mass of the box)

5. Calculate the mass of the box (m):
Weight (W) = mass * acceleration due to gravity (g)
229 N = m * 9.8 m/s^2

Solving for m, we obtain:
m = 229 N / 9.8 m/s^2 = 23.37 kg

6. Substitute the value of the mass into the equation found in step 4:
a = 171.35 N / 23.37 kg = 7.35 m/s^2

7. Apply the equation for uniformly accelerated motion to find the time (t) to cover a distance (d) with initial velocity (v) as zero:
d = (1/2) * a * t^2

Given:
- d = 7 m
- a = 7.35 m/s^2
- v = 0 m/s

Rearranging the equation, we have:
t^2 = (2 * d) / a

Substituting the values, we get:
t^2 = (2 * 7 m) / 7.35 m/s^2
t^2 = 1.92
t ≈ sqrt(1.92)
t ≈ 1.386 s

Therefore, it would take approximately 1.386 seconds to move the box 7 m, starting from rest.