(a) The normal force on the sleigh is 596.36 N.
(b) The magnitude and direction of acceleration of the sleigh is 3.2 m/s² upwards.
The given parameters;
The normal force on the sleigh is calculated as follows;
[tex]F_n = mg \times cos(\theta)\\\\F_n = 63 \times 9.8 \times cos(15)\\\\F_n = 596.36 \ N[/tex]
The magnitude and direction of acceleration of the sleigh is calculated as follows;
[tex]\Sigma F= ma\\\\F - mgsin(\theta) - F_f = ma\\\\F - mgsin(\theta) - \mu F_n = ma\\\\510\ - \ 63 \times 9.8 \times sin15 \ -\ 0.25\times 596.36 = 63a\\\\201 .11 = 63a\\\\a = \frac{201.11}{63} \\\\a = 3.2 \ m/s^2 \ upwards[/tex]
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