We have to work out the accepted value of a roller coaster cart (taking into account friction) when the mass is modified
Here is the answer I got for 56.71 grams:
0=∆EGP+∆EK+Fd
0=mg(hf-hi )+1/2 m(v^2-u^2 )+Fd
0=0.05671 × 9.8 (0.075-0.786)+1/2× 0.05671 (v^2-0^2 )+(0.2184019181 × 1.005505843)
0=-0.395143938+0.028355v^2+0.2196044047
(0.395143938-0.2196044047)/0.028355=v^2
v^2=6.190778815
v=√6.190778815
v=2.488127572
v=2.488m/s
But here is the answer I got for 106.71 grams:
0=∆EGP+∆EK+Fd
0=mg(hf-hi )+1/2 m(v^2-u^2 )+Fd
0=0.10671 × 9.8 (0.075-0.786)+1/2× 0.10671 (v^2-0^2 )+(0.7732989017 × 1.005505843)
0=-0.743533938+0.053355v^2+0.777556564
(0.743533938-0.777556564)/0.053355=v^2
v^2=-0.6353431559
v=√0.6353431559
v=0.7970841586
v=0.797m/s
As the roller coaster has an incline, we were told to work out 'F' by doing: mass × weight × cos45 × mass × weight × tan45
'd' is the distance travelled
'hf' is the final height and 'hi' is the initial height
Can someone please tell me what I have done wrong in these calculations
1 answer