Asked by Christine
Problem::
A 4.80 kg block located on a horizontal frictionless floor is pulled by a cord that exerts a force F=13.5 N at an angle q= 17.5° above the horizontal. What is the speed of the block 3.90 seconds after it starts moving?
What I've got::
Using sin and cos, I've got that Viy is 4.059 and Vix is 12.875.
So, using 12.875 for my initial velocity and the formula Vf=Vi+at, I get Vf=12.875+(2.81)(3.9), which gives me Vf=23.84, but that isn't right. Can someone show me what I'm doing wrong?
A 4.80 kg block located on a horizontal frictionless floor is pulled by a cord that exerts a force F=13.5 N at an angle q= 17.5° above the horizontal. What is the speed of the block 3.90 seconds after it starts moving?
What I've got::
Using sin and cos, I've got that Viy is 4.059 and Vix is 12.875.
So, using 12.875 for my initial velocity and the formula Vf=Vi+at, I get Vf=12.875+(2.81)(3.9), which gives me Vf=23.84, but that isn't right. Can someone show me what I'm doing wrong?
Answers
Answered by
bobpursley
Vi is zero.
12.875 is the pulling force
a=pullingforce/m=12.875/4.80=your acceleration
Vf=a*3.9seconds
12.875 is the pulling force
a=pullingforce/m=12.875/4.80=your acceleration
Vf=a*3.9seconds
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