What is the equation to find the Average speed

A 6 kg box with initial speed 5 m/s slides across the floor and comes to a stop after 1.9 s. What is the coefficient of kinetic

Ilia_Sergeevich [38]

Answer:

$\mu_k=0.27$

Explanation:

According to the free body diagram, in this case, we have:

$\sum F_x:-F_k=ma\\\sum F_y:N=mg$

Recall that the force of friction is given by:

$F_k=\mu_k N$

Replacing and solving for the coefficient of kinetic friction:

$-\mu_kN=ma\\-\mu_k(mg)=ma\\\mu_k=-\frac{a}{g}$

We have an uniformly accelerated motion. Thus, the acceleration is defined as:

$a=\frac{v_f-v_0}{t}\\a=\frac{0-5\frac{m}{s}}{1.9s}\\a=-2.63\frac{m}{s^2}$

Finally, we calculate $\mu_k$ :

$\mu_k=-\frac{-2.63\frac{m}{s^2}}{9.8\frac{m}{s^2}}\\\mu_k=0.27$

4 0

3 years ago

A child is riding in a child-restraint chair, securely fastened to the seat of a car. Assume the car has speed 47 km/h when it h

Elina [12.6K]

Answer: F = 1235 N

Explanation: Newton's Second Law of Motion describes the effect of mass and net force upon acceleration: $F_{net}=m.a$

Acceleration is the change of velocity in a period of time: $a=\frac{\Delta v}{\Delta t}$

Velocity of the car is in km/h. Transforming it in m/s:

$v=\frac{47.10^{3}}{36.10^{2}}$

v = 13 m/s

At the moment the car decelerates, acceleration is

$a=\frac{13}{0.2}$

a = 65 m/s²

Then, force will be

$F_{net}=19(65)$

$F_{net}$ = 1235 N

The horizontal net force the straps of the restraint chair exerted on the child to hold her is 1235 newtons.

5 0

2 years ago

Please, I need help with this.

solmaris [256]

this is the answer :)

6 0

3 years ago

Read 2 more answers

Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 10

Tema [17]

Answer:

Explanation:

(a) It is given that Joseph jogs on a straight road of 300m in a time interval of 2 minutes and 30 seconds, which is equal to 150seconds. Therefore, when Joseph jogs from point A to point B, he covers a distance of 300m in time of 150seconds. Hence, his average speed is 300m/150s=2ms^−1. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m.

Hence, his average velocity is 300m/150s=2ms^−1

(b) Then it is given that he turns back and points B and jogs on the same road but in the opposite direction for a time interval for 1 minute and covers a distance of 100m.If we consider the whole motion of Joseph, i.e. from point A to point C, then he covers a total distance of 300m+100m=400m. And he covers this total distance in a time interval of 2.5min+1min=3.5min=210s.

Therefore, his average speed for this journey is 400m210s=1.9ms−1.

For the same journey is displacement is equal to the distance between the points A and C,i.e. 300m−100m=200m.

Hence, his average velocity for this case is 200m/210s=0.95ms^−1

7 0

3 years ago

1.Suppose someone pulls a cart up a ramp a distance of 85cm along the ramp with a force of 15N.

Drupady [299]

1. 12.75 J

Assuming that the force applied is parallel to the ramp, so it is parallel to the displacement of the cart, the work done by the force is

$W=Fd$

where

F = 15 N is the magnitude of the force

d = 85 cm = 0.85 m is the displacement of the cart

Substituting in the formula, we get

$W=(15 N)(0.85 m)=12.75 J$

2. 10.6 N

In this part, the cart reaches the same vertical height as in part A. This means that the same work has been done (because the work done is equal to the gain in gravitational potential energy of the object: but if the vertical height reached is the same, then the gain in gravitational potential energy is the same, so the work done must be the same).

Therefore, the work done is

$W=Fd=12.75 J$

However, in this case the displacement is

d = 120 cm = 1.20 m

Therefore, the magnitude of the force in this case is

$F=\frac{W}{d}=\frac{12.75 J}{1.20 m}=10.6 N$

3 0

2 years ago