6. A sled traveling at a speed of 3.0 m/s slows to a stop 4.0 m from the point where its passenger rolled off. What is the magnitude of the horizontal net force that
slows the 110 N sled? (Assume ag = 10m/s2)
D

The magnitude of the horizontal net force acting on the sled is 12.375 N.

Given data:

The initial speed of sled is, u = 3.0 m/s.

The final speed of sled is, v = 0 m/s.

The distance covered before stopping is, s = 4.0 m.

The weight of sled is, W = 110 N.

Using the Newton's second law, the net horizontal force acting on the sled is given as,

F = ma

here,

m is the mass of sled.

a is the acceleration of sled.

Mass is calculated from the weight as,

W = mg

110 = m (10)

m = 11 kg.

Now, using the second kinematic equation of motion to obtain the acceleration of sled as,

Then the magnitude of net horizontal force is,

Thus, we can conclude that the magnitude of the horizontal net force is 12.375 N.

Learn more about the Newton's second law here:

link

-12.38 N

Explanation:

This question requires you to first find the acceleration then apply Newton's 2nd law of motion to find the force.

To find deceleration use the formula;

v²= u²+2ad ---where

v= final speed = 0 m/s

u= initial speed = 3.0 m/s

a= acceleration { in this case , deceleration} = ?

d= distance from the point where its passenger rolled off= 4.0 m

therefore ;

Mass of the sled is calculated from its weight force.

Weight force = mg

mass of sled = weight force / g  where g= 10 m/s²

mass of sled = 110/10 = 11kg

The magnitude of horizontal net force will be : F= ma  where ;

m= mass of sled = 11 kg

a= deceleration = -9/8 m/s²

F = 11 * -9/8 = -12.38 N

Force is acting on the opposite direction of the initial motion of the sled.

1) kg

2) m

3) g

4) mL

5) mm

6) L

7) km

8) cm

9) mg

10) 2g

11) 104000m

12) 4.8m

13) 5600g

14) 8cm

15) 5000mL

16) 0.198kg

17) 0.075L

18) 0.15m

19) 560cm

20) 16mm

21) 25km

22) 6500mg

23) 66mm

24) 0.12g