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Anika [276]
3 years ago
12

A stuntman drives a car with a mass of 1500 kg on a drawbridge. The car accelerates with a constant force of 10,000 N. While he

is driving, the drawbridge is raised to an incline of 30°. What is the car's new acceleration on this incline, ignoring the force due to air resistance? (recall that g=9.8 m/s^2)
Physics
1 answer:
Viktor [21]3 years ago
7 0

Answer:

1.77 m/s^2

Explanation:

There are two forces acting on the car along the direction parallel to the incline:

- The driving force of 10,000 N, which pushes forward

- The component of the weigth of the car parallel to the incline, which pulls backward

The component of the weight of the car parallel to the incline is:

W_p = mg sin \theta=(1500 kg)(9.8 m/s^2)( sin 30^{\circ})=7350 N

So now we can apply Newton's second law to find the acceleration of the car:

F-W_p = ma\\a=\frac{F-W_p}{m}=\frac{10000 N-7350 N}{1500 kg}=1.77 m/s^2

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It takes Seymour, a slimy slug, 20 minutes to travel from his favorite bush to the local trash can (a distance of 30 meters), ho
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A 2N and 6N force pull on an object to the right and a 4N force pulls to the left a 0.5kg object. What is the net force on the o
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Explanation:

F net = 2+6-4 ( 2 and 6 N are in same direction so they get added, 4N in opposite direction so it will be subtracted)

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3 years ago
A 4.25 kg block is sent up a ramp inclined at an angle theta=37.5° from the horizontal. It is given an initial velocity ????0=15
wel

Answer:

d = 11.79 m

Explanation:

Known data

m=4.25 kg  : mass of the block

θ =37.5°  :angle θ of the ramp with respect to the horizontal direction

μk= 0.460  : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the block

W: Weight of the block : In vertical direction

N : Normal force : perpendicular to the ramp

f : Friction force: parallel to the ramp

Calculated of the W

W= m*g

W=  4.25 kg* 9.8 m/s² = 41.65 N

x-y weight components

Wx= Wsin θ= 41.65*sin 37.5° = 25.35 N

Wy= Wcos θ =41.65*cos 37.5° =33.04 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay    ay = 0

N - Wy = 0

N = Wy

N = 33.04 N

Calculated of the f

f = μk* N= 0.460*33.04

f = 15.2 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

-Wx-f = m*a

 -25.35-15.2 = (4.25)*a

-40.55 =  (4.25)*a

a = (-40.55)/ (4.25)

a = -9.54 m/s²

Kinematics of the block

Because the block moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d Formula (2)

Where:  

d:displacement  (m)

v₀: initial speed  (m/s)

vf: final speed   (m/s)

Data:

v₀ = 15 m/s

vf = 0

a = -9.54 m/s²

We replace data in the formula (2)  to calculate the distance along the ramp the block reaches before stopping (d)

vf²=v₀²+2*a*d

0 = (15)²+2*(-9.54)*d

2*(9.54)*d =   (15)²

(19.08)*d = 225

d = 225 / (19.08)

d = 11.79 m

3 0
3 years ago
A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 joules (with the poten
lions [1.4K]

Answer:

 h = 20 m

Explanation:

given.

height, h = 10 m

Potential energy at 10 m = 50 J

Kinetic energy at 10 m = 50 J

maximum height the ball will reach, H = ?

Total energy of the system

T E = 50 J + 50 J

T E = 100 J

now,

A h = 10 m

P E = m g h

50 = m g x 10

mg = 5 ..............(1)

at the top most Point the only Potential energy will be acting on the body.

now, TE = Potential energy

 100 = m g h

5 h = 100

 h = 20 m

hence, the maximum height reached by the ball is equal to 20 m.

5 0
3 years ago
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