225k views
0 votes
A 2.97 kg block initially at rest is pulled to the right along a horizontal surface by a constant, horizontal force of 12.9 N. The coefficient of kinetic friction is 0.126. The acceleration of gravity is 9.8 m/s 2 . Find the speed of the block after it has moved 2.92 m. Answer in units of m/s.

User Lucha
by
7.7k points

2 Answers

2 votes

Final answer:

The block accelerates at 3.1 m/s² due to the net force after overcoming kinetic friction and reaches a speed of approximately 4.25 m/s after moving 2.92 m on the surface.

Step-by-step explanation:

To calculate the speed of a 2.97 kg block after it has moved 2.92 m on a horizontal surface, we need to apply the principles of mechanics. We need to consider the net force acting on the block, which is the difference between the applied force and the frictional force. The frictional force is given by the product of the coefficient of kinetic friction and the normal force, which, because the surface is horizontal, equals the weight of the block due to gravity. The net force will then allow us to find the acceleration of the block using Newton's second law (F = ma), and finally, we can determine the block's speed using kinematic equations for uniformly accelerated motion.

First, calculate the frictional force (f k) as:

  • f k = μk ⋅ N
  • f k = 0.126 ⋅ (2.97 kg ⋅ 9.8 m/s2)
  • f k = 3.68844 N

Next, calculate the net force (Fnet) acting on the block:

  • Fnet = applied force - frictional force
  • Fnet = 12.9 N - 3.68844 N
  • Fnet = 9.21156 N

Using Newton's second law, calculate the acceleration (a):

  • a = Fnet / m
  • a = 9.21156 N / 2.97 kg
  • a = 3.1 m/s2

Finally, use the kinematic equation to find the final speed (v) after travelling 2.92 m:

  • v2 = u2 + 2 ⋅ a ⋅ d
  • v2 = 0 + 2 ⋅ 3.1 m/s2 ⋅ 2.92 m
  • v2 = 18.092 m2/s2
  • v = √(18.092 m2/s2)
  • v ≈ 4.25 m/s

Therefore, the speed of the block after it has moved 2.92 m is approximately 4.25 m/s.

User Bonh
by
8.5k points
6 votes

Answer:

The speed of the block after it has moved 2.92 m is 4.26 m/s

Step-by-step explanation:

Given;

mass of the block, m = 2.97 kg

initial velocity of the block, u = 0

applied horizontal force, Fx = 12.9 N

Coefficient of static friction, μ = 0.126

The frictional force is given by;


F_s = \mu R\\\\F_s = \mu (mg)\\\\F_s = 0.126 (2.97*9.8)\\\\F_s = 3.667 \ N

The net horizontal force is given by;


F_(net) = F_x - F_s\\\\F_(net) = 12.9 - 3.667\\\\F_(net) = 9.233 \ N

The acceleration of the block is given by;


a = (F_(net))/(m)\\\\a = (9.233)/(2.97)\\\\a = 3.11 \ m/s^2

The speed of the block after it has moved 2.92 m is given by the following kinematic equation;

v² = u² + 2as

v² = 0 + 2as

v² = 2as

v = √2as

v = √(2 x 3.11 x 2.92)

v = 4.26 m/s

Therefore, the speed of the block after it has moved 2.92 m is 4.26 m/s

User Dore
by
8.9k points