Nourwaa2003idn
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2. You push a 12.3-kg shopping cart with a force of 10.1 N.
(a) What is the acceleration of the cart?
(b) If the cart starts from rest, how far does it move in 2.50 s?


Plz answer quick

Sagot :

Answer:

Explanation:

The important part about this problem is the acceleration, which you must find by using

F = ma,

where F is the force applied,

m is the mass,

and a is the acceleration.

F = ma

10.1 N = (12.3 kg)a

a = 0.8211 m/s^2

Now use the kinematics equations.

d = vot + (1/2)at^2

where d is the distance moved,

vo is the initial velocity (0),

a is the acceleration,

and t is the time.

d = vot + (1/2)at^2

d = (0 m/s)(2.5 s) + (1/2)(0.8211 m/s^2)(2.5 s)^2

d = 2.57 m

Lanuelidn

a. The acceleration of the cart, is 0.821 [tex]m/s^2[/tex].

b. The car moved a distance of 30.63 meters starting from rest.

Given the following data:

  • Mass of shopping cart = 12.3 kg
  • Force = 10.1 N
  • Initial velocity, u = 0 m/s (since the cart is starting from rest).
  • Time, t = 2.50 seconds.

a. To find the acceleration of the cart, we would apply Newton's Second Law of Motion:

Mathematically, Newton's Second Law of Motion is given by this formula;

[tex]Acceleration = \frac{Force}{Mass}[/tex]

Substituting the given values, we have;

[tex]Acceleration = \frac{10.1}{12.3}[/tex]

Acceleration = 0.821 [tex]m/s^2[/tex]

Therefore, the acceleration of the cart, is 0.821 [tex]m/s^2[/tex].

b. To find the distance covered by the shopping cart, we would use the second equation of motion.

Mathematically, the second equation of motion is given by the formula;

[tex]S = ut + \frac{1}{2} at^2[/tex]

Where:

  • S is the distance covered.
  • u is the initial velocity.
  • a is the acceleration.
  • t is the time measured in seconds.

Substituting the values into the formula, we have;

[tex]S = 0(2.5) + \frac{1}{2} (9.8)(2.5)^2\\\\S = 0 + 4.9(6.25)[/tex]

Distance, S = 30.63 meters.

Therefore, the car moved a distance of 30.63 meters starting from rest.

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