An object falls from rest at a rate of 10 m/s / s . How fast was it traveling after 5 seconds of fall?

While a roofer is working on a roof that slants at 37.0 ∘∘ above the horizontal, he accidentally nudges his 92.0 NN toolbox, cau

eduard

Answer:

The speed is $v =8.17 m/s$

Explanation:

From the question we are told that

The angle of slant is $\theta = 37.0^o$

The weight of the toolbox is $W_t = 92.0N$

The mass of the toolbox is $m = \frac{92}{9.8} = 9.286kg$

The start point is $d = 4.25m$ from lower edge of roof

The kinetic frictional force is $F_f = 22.0N$

Generally the net work done on this tool box can be mathematically represented as

$Net \ work done = Workdone \ due \ to \ Weight + Workdone \ due \ to \ Friction$

The workdone due to weigh is = $mgsin \theta * d$

The workdone due to friction is = $F_f \ cos\theta * d$

Substituting this into the equation for net workdone

$W_{net} = mgsin\theta * d + F_f \ cos \theta *d$

Substituting values

$W_{net} = 92 * sin (37) * 4.25 + 22 cos (37) * 4.25$

$= 309.98 J$

According to work energy theorem

$W_{net} = \Delta Kinetic \ Energy$

$W_{net} = \frac{1}{2} m (v - u)^2$

From the question we are told that it started from rest so u = 0 m/s

$W_{net} = \frac{1}{2} * m v^2$

Making v the subject

$v = \sqrt{\frac{2 W_{net}}{m} }$

Substituting value

$v = \sqrt{\frac{2 * 309.98}{9.286} }$

$v =8.17 m/s$

6 0

3 years ago

when in object moves stop moving changes speed or changes direction how do scientist describe that condition?

Kruka [31]

im taking test rn nd this question was on there, im saying unbalanced force, if it not correct i will put right answer, but im pretty sure the answer is unblanced

6 0

3 years ago

Read 2 more answers

A 1.2kg ball rolls forward with an acceleration of 1.11 m/s. What is the net force on the ball

grandymaker [24]

Answer:

1.332 N

Explanation:

Net Force = Mass x Acceleration
1.2 x 1.11 = 1.332 N

I'm so sorry if I'm wrong.

8 0

1 year ago

Um comentarista de futebol certa vez comentou:"A bola bateu na trave e voltou duas vezes mais forte". Sabendo que quando a bola

ryzh [129]

Answer:

Por ela ter batido na trave, não tem como voltar 2x mais forte, por que toda ação correspondente a uma reação de igual intensidade, mas que atua no sentido oposto

Explanation:

7 0

2 years ago

A car of mass 1000 kg is moving at 25 m/s. It collides with a car of mass 1200 kg moving at 30 m/s. When the cars collide, they

Alinara [238K]

Answer:

The total momentum of the cars before the collision is 61,000 kg.m/s

The total momentum of the cars after the collision is 61,000 kg.m/s

The velocity of the cars after the collision is 27.727 m/s

Explanation:

Given;

mass of the first car, m₁ = 1000 kg

initial velocity of the car, u₁ = 25 m/s

mass of the second car, m₂ = 1200 kg

initial velocity of the second car, u₂ = 30 m/s

The common velocity of the cars after collision = v

The total momentum of the cars before collision is calculated as;

P₁ = m₁u₁ + m₂u₂

P₁ = (1000 x 25) + (1200 x 30)

P₁ = 61,000 kg.m/s

The total momentum of the cars after collision is calculated as;

P₂ = m₁v + m₂v

where;

v is the common velocities of the cars after collision since they stick together.

P₂ = v(m₁ + m₂)

To determine "v" apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁ + m₂u₂ = v(m₁ + m₂)

(1000 x 25) + (1200 x 30) = v(1000 + 1200)

61,000 = 2,200v

v = 61,000/2,200

v = 27.727 m/s

The total momentum after collsion = v(m₁ + m₂)

= 27.727(1000 + 1200)

= 61,000 kg.m/s

Thus, momentum before and after collsion are equal.

8 0

2 years ago