Recent content by a18c18

Okay thank you. so if I use the formula Ki+Ui=Kf+Uf I don't need to multiply gravity by cos(angle)?

So I don't need to account for the angle any where in the problem? Other than the acceleration is there anything else wrong my work?

Okay thank you. I tried a new approach using the conservation of energy and the formula Ki+Ui=Kf+Uf. This would be 1/2mVi^2+mgh=1/2mVf^2+mgh. 1/2mVi^2+m(9.81 cos 51)*0=1/2*0^2+m(9.81 cos 51)*26 1/2mVi^2=m(9.81 cos 51)*26 So after the masses cancel out, Vi should be 17.9 m/s Did I do this correctly?

Okay so the vertical component of acceleration would be 9.81*cos51 right? And then plug this into v2=vo2+2ax using 26 for x? And how do I get the answer from the vertical and horizontal components of the initial speed?

Homework Statement A stone is thrown upward at an angle of 51° above the horizontal. Its maximum height during the trajectory is 26 m. What was the stone's initial speed? m/s Homework Equations a=v2/r v2=vo2+2ax C=2\pir The Attempt at a Solution I figured the acceleration is...

Oh okay so it was sin instead of cos thank you very much!

Since it is a frictionless surface I didn't think there were any forces other than normal and gravity?

Homework Statement A 5.0 kg block slides along a frictionless horizontal surface with a speed of 7.0 m/s. After sliding a distance of 3.0 m, the block makes a smooth transition to a frictionless ramp inclined at an angle of 40° to the horizontal. How far up the ramp does the block slide...

Okay i got it. Thank you so much!

Homework Statement You run a race with your friend. At first you each have the same kinetic energy, but then you find that she is beating you. When you increase your speed by 21%, you are running at the same speed she is. If your mass is 77 kg, what is her mass? Homework Equations...

Okay thank you! i found the accelerations using v=d/t and then a=change in v/t. But now I am trying to plug it into a=F(net)/m and don't see how to find this without knowing either u(kinetic) or mass. The sum of the forces would be F(normal)+F(kinetic)+tension F(normal)=mg...

Okay i retried this and got: x: f(kinetic)+100*cos30=mg*cos15 y:F(normal)=mg*cos(15) I got that F(normal)=202.84 then I plugged this into f(kinetic)+100*cos30=mg*cos15 which would be f(kinetic)=202.84-100cos30 which would be 116.24 but this answer was incorrect Does anyone know why?

I did calculate the accelerations actually but I wasn't sure what to do with that. I used v^2=v(initial)^2-2ad and got that the first acceleration is 1.219 and the second is .473

Homework Statement A mass m1 on a horizontal shelf is attached by a thin string that passes over a frictionless pulley to a 2.8 kg mass (m2) that hangs over the side of the shelf 1.6 m above the ground. The system is released from rest at t = 0 and the 2.8 kg mass strikes the ground at t =...

Homework Statement A sled weighing 210 N rests on a 15° incline, held in place by static friction. The coefficient of static friction is 0.50. (c) The sled is now pulled up the incline at constant speed by a child. The child weighs 546 N and pulls on the rope with a constant force of 100...