A 4.30 kg block located on a horizontal frictionless floor is pulled by a cord that exerts a force F=11.90 N at an angle theta=19.5o above the horizontal, as shown. What is the speed of the block 5.90 seconds after it starts moving?
I know you're supposed to treat x and y seperately so I found the adjacent side of the tension which was 11.9 cos 19.5 = 11.2, and the opposite side which was 4.02. 4.30kg*9.8 = 42.14 N as the Normal force/gravity.
y = 42.24 - 4.02 = 38.12 N
x = 11.12 N?
Fnet = ma so (38.12+11.12) = 4.3(a)
I don't see how this could turn into speed unless I found the acceleration from this, which is too high and multiplied it by the seconds (5.9) to get m/s. please help!
2007-02-12 19:05:53 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The block is on a horizontal surface with no friction. Therefore the weight of the block (m*g) has no effect on this situation, since the block is not going to move vertically. Only the horizontal (x) component of force will result in motion. You have calculated that properly as 11.2N. You do not need the y component. The horizontal acceleration is 11.2N divided by the mass. Velocity is accel x time.
2007-02-12 19:21:29 · answer #1 · answered by gp4rts 7 · 0⤊ 0⤋
dont forget that you would divide the acceleration by seconds to get speed. Acceleration is measured in m/s^2.
And did you make sure the force of gravity was pointing down? Sounds crazy I know, but the second rule of physics is that you can't push a rope. The mass of the block is in newtons and pointing down. The force of the rope is as you say. The m/s^2 get into the equation from the gravity.
2007-02-12 19:17:14 · answer #2 · answered by Anonymous · 0⤊ 0⤋
both pulley kit you've shown is fairly no diverse than a unmarried pulley, in which case the pressure in the string would actual be the mass of between the 500g weights, or 5 N.
2016-11-27 19:42:48 · answer #3 · answered by Anonymous · 0⤊ 0⤋