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Mrrafil [7]
3 years ago
14

A toy cart is pulled a distance of 6.00 m in a straight line across the floor. The force pulling the cart has a magnitude of 20.

0 N and is directed at 37° above the horizontal. What is the work done by this force?

Physics
2 answers:
aivan3 [116]3 years ago
8 0

Answer:

W = 95.8J

Explanation:

Given

S = 6.00m

F =20.0N

Theta = 37°

W = FSCos(theta)

W = 20×6.0×Cos(37)

W = 95.8J

Novosadov [1.4K]3 years ago
6 0

Given Information:  

Horizontal distance = d = 6.0 m  

Force = F = 20 N

Angle = θ = 37°

Required Information:  

Work done = W = ?

Answer:  

Work done = 95.84 Joules

Explanation:  

The work done on toy cart is given by

W = Fdcos(θ)

Where F is the force applied, d is the distance by which the toy cart has been displaced, and θ is the angle between F and horizontal floor.  

As you can see in the attached diagram, The force pulling the cart makes an angle of 37 with respect to the horizontal surface.

W = 20*6*cos(37°)

W = 95.84 Joules

Therefore, 95.84 Joules of work has been by a force of 20 N to displace the toy cart by 6 meters.

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Newton's second law and the kinematic relations allow to find the results for the questions about forces and the movement of the block are:

    B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º

    C) The maximum force is: F = 24 N

    D) The time to stop the block is: t = 25 s

 

Newton's second law establishes a relationship between the net force, the mass, and the acceleration of the body. In the special case that the acceleration is zero it is called the equilibrium condition.

B) They indicate a diagram of forces on the block, let's look for the components of the force that the block maintains with zero acceleration, in the attached we have a free-body diagram including the force applied to keep the system in equilibrium.

x-axis

      -10 + 12 sin 60 + Fₓ = 0

        Fₓ = 10- 12 sin 60 = -0.39 N

y-axis

       12 cos 60 - 6 + F_y = 0

        F_y = 6 - 12 cos 60 = 0 N

We can give the result of the force in two ways:

  • Form of coordinates F = -0.39 i ^ N
  • Form of module and angle.

Let's use Pythagoras' theorem to find the modulus.

       F = \sqrt{F_x^2 + F_y^2 } \\F = \sqrt{0.39^2 +0^2}  

       F = 0.39N

We use trigonometry for the angle.

       tan \theta = \frac{F_y}{F_x}

       tan θ=  0º

The component of the force is negative therefore this angle is in the second quadrant, to measure the angle from the positive side of the x axis in a counterclockwise direction.

        θ = 180 + θ'

        θ = 180 + 0

        θ = 180º

C) if the three forces can be moved and the maximum force occurs when they are all linear.

          10+ 6 + 6 + F = 0

          F = -24 N

D) if we maintain this force and eliminate the other three, the block stops, let's look for its acceleration.

          a = \frac{F}{m}  

          a = \frac{24}{6}  

         a =  4 m / s²

The acceleration is in the opposite direction of the initial velocity of the block v₀ = -100 m / s

If we use kinematic relations.

        v = v₀ - a t

Final velocity when stopped is zero

         t = \frac{0-v_o}{a}

         t = 100/4

         t = 25 s

In conclusion using Newton's second law and the kinematics relations we can find the results for the questions about the forces and the motion of the block are:

    B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º

    C) The maximum force is: F = 24 N

    D) The time to stop the block is: t = 25 s

Learn more about Newton's second law here: brainly.com/question/25545050

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