PLZ help !!!!!!!!!!!!!!! its due today

Batman is driving the Batmobile on the road in Gotham City. Which of the following may be true about this ride?

Basile [38]

Answer:

A

Explanation:

i dont really have an explanation but it just seems correct yk?

7 0

3 years ago

A 62.6-kg skateboarder starts out with a speed of 2.18 m/s. He does 113 J of work on himself by pushing with his feet against th

n200080 [17]

Answer:

a. Wgra=786.09J

b. 1.28m

Explanation:

The change in the potential energy is the work done by the gravitational force.

For this problem you have to take into account that the total work done is given by the change in the kinetic energy

$W_{tot}=\Delta E_k=\frac{m}{2}(v_f^2-v_0^2)\\W_{tot}=\frac{62.6kg}{2}((6.55\frac{m}{s})^2-(2.18\frac{m}{s})^2)=1194.09J$

Furthermore the total work is the contribution of the work done by the skater, the gravitational force and the friction

$W_{tot}=W_{ska}+W_{fric}+W_{gra}$

(a) by separating Wfric you have

$W_{gra}=W_{tot}-W_{fric}-W_{ska}=1194.09J-295J-113J=786.09J$

(b) It is only necessary to use the expression for the work done by gravitational force

$W_{grav}=mgh\\h=\frac{W_{grav}}{mg}=\frac{786.09J}{(62.6kg)(9.8\frac{m}{s^2})}=1.28m$

HOPE THIS HELPS!!

6 0

3 years ago

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

rjkz [21]

Answer:

a) $F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

b) $\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

Explanation:

In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).

After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:

$\Sigma F_{y}=0$

We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:

$-F_{y}-W+N=0$

When solving for the normal force we get:

$N=F_{y}+W$

we know that

W=mg

and

$F_{y}=Fsin \theta$

so after substituting we get that

N=F sin θ +mg

We also know that the kinetic friction is defined to be:

$f_{k}=\mu_{k}N$

so we can find the kinetic friction by substituting for N, so we get:

$f_{k}=\mu_{k}(F sin \theta +mg)$

Now we can find the sum of forces in x:

$\Sigma F_{x}=0$

so after analyzing the diagram we can build our sum of forces to be:

$-f+F_{x}=0$

we know that:

$F_{x}=Fcos \theta$

so we can substitute the equations we already have in the sum of forces on x so we get:

$-\mu_{k}(F sin \theta +mg)+Fcos \theta=0$

so now we can solve for the force, we start by distributing $\mu_{k}$ so we get:

$-\mu_{k}F sin \theta -\mu_{k}mg)+Fcos \theta=0$

we add $\mu_{k}mg$ to both sides so we get:

$-\mu_{k}F sin \theta +Fcos \theta=\mu_{k}mg$

Nos we factor F so we get:

$F(cos \theta-\mu_{k} sin \theta)=\mu_{k}mg$

and now we divide both sides of the equation into $(cos \theta-\mu_{k} sin \theta)$ so we get:

$F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

which is our answer to part a.

Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:

$f_{s}=\mu_{s}(F sin \theta +mg)$